First, I am going to provide a little background on equivalent pressures at different altitudes from Earth's surface.
Layers of Earth's Atmosphere
Troposphere to Mesosphere
At sea level, the neutral atmosphere of Earth has a pressure of ~$10^{5}$ Pa (or ~1000 mbars).
The below image from https://en.wikipedia.org/wiki/File:Comparison_US_standard_atmosphere_1962.svg shows the broad range of temperatures/pressures of Earth's atmosphere.
Ionosphere
The region where the atmosphere transitions from mostly neutral to mostly ionized gas (called a plasma) is called the ionosphere. The altitudes defining this region vary (due to solar variability), but are generally defined as ranging from ~60-1000 km. The free electron number density in the ionosphere varies greatly from ~$10^{3} - 10^{6}$ # $cm^{-3}$ (or number of particles per cubic centimeter). The temperature varies from few 100 K to ~1500 K. Thus, if we treated it like an ideal gas the thermal pressure of the charged particles would range from a few $10^{-12}$ Pa to few $10^{-8}$ Pa.
Thus, the ratio of sea level pressure to the plasma constituents would be ~$10^{13} - 10^{17}$.
Plasmasphere
The region immediately surrounding the ionosphere is called the plasmasphere, which can extend to altitudes as low as a few $R_{E}$ up to ~6 $R_{E}$. The density ranges from several 100 # $cm^{-3}$ down to ~10 # $cm^{-3}$ and temperatures vary greatly, from ~6000-35,000 K. Again, these ranges correspond to thermal pressures of $10^{-13}$ Pa to few $10^{-11}$ Pa.
Thus, the ratio of sea level pressure to the plasma constituents would be ~$10^{16} - 10^{18}$.
Outer Magnetosphere
The "best" vacuum that we can easily access is the Earth's outer magnetosphere, which has a density ranging from ~0.01-1.0 particles per cubic centimeter. The temperatures in the outer magnetosphere can vary greatly from ~$10^{5}$ K to greater than $10^{9}$ K (i.e., if one converts radiation belt particle energies, e.g., 100s of keV, to an equivalent temperature). Thus, the range of equivalent ideal gas thermal pressures would be few $10^{-14}$ Pa to a few $10^{-8}$ Pa.
Thus, the ratio of sea level pressure to the plasma constituents would be ~$10^{13} - 10^{19}$.
Solar Wind
The solar wind - the supersonic flow of plasma from the sun's upper atmosphere - has densities and temperatures ranging from ~0.5-50 # $cm^{-3}$ and ~$10^{4}$ K up to ~$10^{6}$ K, respectively (for references, see https://physics.stackexchange.com/a/179057/59023). Thus, the range of equivalent ideal gas thermal pressures would be few ~$10^{-14} - 10^{-10}$ Pa.
Thus, the ratio of sea level pressure to the plasma constituents would be ~$10^{15} - 10^{19}$.
Answers
At what altitude would the air be too thin to carry a sound wave?
Short Answer
For all practical purposes, there are no regions of space completely devoid of some kind of sound.
Long/Detailed Answer
Interestingly, one can have the traditional sound wave (i.e., a longitudinal oscillation mediated by gas particle collisions) propagate into the upper atmosphere and even into the ionosphere. The point where such a sound wave would experience strong damping is where the collisional mean free path becomes too large to support the oscillations, i.e., this would occur when the average time between collisions becomes comparable to the wave frequency. Thus, the oscillations would have no restoring force and would damp out (for an electromagnetic analogy, see evanescent waves).
It is a common belief that in space there is no sound...?
No, there are sound waves that start in space can propagate in almost zero pressures, well, less than $10^{-14}$ Pa is as close to vacuum as one needs since the best vacuums we can produce in a lab is ~$10^{-11}$ Pa.
There are multiple sound waves in space, including ion acoustic waves and magnetosonic waves. Ion acoustic waves have been seen all the way out at Saturn, where the solar wind density and temperature can be as low as ~0.1 # $cm^{-3}$ and less than ~$10^{4}$ K, or thermal pressures below $10^{-14}$ Pa (note that the ram or dynamic pressure is generally ~2-4 orders of magnitude higher owing to the high speed of the solar wind). These modes have been observed as far out as Neptune, throughout the solar wind, and as in close as ~0.3 AU.
There is no reason to expect that such modes would not exist in the interstellar medium, where densities and temperatures can be as low as ~0.1 # $cm^{-3}$ and ~$10^{3}$ K, corresponding to $10^{-15}$ Pa.
The intra galaxy cluster medium is even more tenuous but much hotter, with densities and temperatures as low as ~$10^{-4}$ # $cm^{-3}$ and ~$10^{7}$ K, corresponding to $10^{-14}$ Pa (e.g., see arXiv e-print 1406.4410). Again, the ubiquity of ion acoustic waves in the interplanetary medium suggests we should expect these in nearly all regions of space.
Update/Edit
Let me ask the question in a slightly different manner.
At what altitude would I no longer be able to hear a 100 dB source (ignoring suffication)?
The intensity of sound decreases as $I\left( r \right) \propto r^{-2}$ while sound pressure decreases as $P\left( r \right) \propto r^{-1}$. The hearing threshold is a function of frequency, because the human ear does not have a flat frequency response, but it is generally accepted as being ~20 $\mu$Pa for 1 atmosphere and 25$^{\circ} C$ at 1000 Hz. The sound pressure level (measured in dB) is given by:
$$
L_{p}\left( r \right) = 20 \ \log_{10} \left( \frac{ P\left( r \right) }{ P_{o} } \right)
$$
where we set $P_{o}$ ~ 20 $\mu$Pa. Then a 100 dB source corresponds to ~2 Pa at the source. This would drop to $P_{o}$ at a distance of ~$10^{5}$ m, ignoring any acoustic impedance or losses and assuming the pressure and temperature are the same as the reference parameters for $P_{o}$.
The reference sound intensity, $I_{o}$, depends upon the characteristic acoustic impedance, $z_{o}$, as $I_{o} = P_{o}^{2}/z_{o}$. We know that $z_{o} = \rho \ C_{s}$, where $\rho$ is the mass density and $C_{s}$ is the speed of sound. We can model $\rho = \rho \left( h \right)$ using a known atmospheric scale height and an exponential decay (which reproduces the orange line the figure above) and take a set of values from the blue in the figure above for $C_{s}$ (see table below). Then we find that $z_{o}$ ranges from ~0.003-416 Pa s/m from 0-100 km altitude. If we use the human hearing threshold for $P_{o}$, then $I_{o}$ ranges from ~$10^{-13} - 10^{-7}$ W $m^{2}$.
Altitude [km] | Cs [m/s] | Density [kg/m^3] | z_o [Pa s/m] | I_o [W/m^2]
-------------------------------------------------------------------------
10 | 300.0 | 3.736e-01 | 1.121e+02 | 3.569e-12
20 | 295.0 | 1.152e-01 | 3.399e+01 | 1.177e-11
30 | 301.3 | 3.553e-02 | 1.071e+01 | 3.736e-11
40 | 316.0 | 1.096e-02 | 3.462e+00 | 1.155e-10
50 | 329.0 | 3.378e-03 | 1.112e+00 | 3.599e-10
60 | 320.0 | 1.042e-03 | 3.334e-01 | 1.200e-09
70 | 298.0 | 3.213e-04 | 9.573e-02 | 4.178e-09
80 | 276.0 | 9.906e-05 | 2.735e-02 | 1.463e-08
90 | 276.0 | 3.055e-05 | 8.431e-03 | 4.744e-08
100 | 287.0 | 9.420e-06 | 2.703e-03 | 1.480e-07
Since $I_{o}$ increases as we increase altitude, then the intensity of our source would have to increase as well to maintain its initial $L_{o}$ = 100 dB level (i.e., $I_{src}\left( h \right) = I_{o}\left( h \right) 10^{L_{o}/10}$). The intensity at the source, $I_{src}$, then ranges from ~0.01-1500 W $m^{2}$.
Let's assume we use the same intensity at sea level and bring the speaker up in altitude, then the intensity level at the source drops with increasing altitude as:
$$
L_{i,src}\left( h \right) = 10 \ \log_{10} \left( \frac{ I_{src}\left( 0 \right) }{ I_{o}\left( h \right) } \right)
$$
Then $L_{i,src}$ varies from 100 dB at sea level to ~94 dB by 10 km, ~79 dB by 50 km, and ~48 dB by 100 km.
We estimate the intensity level at a given distance away from the source as:
$$
L_{r}\left( h, r \right) = L_{i,src}\left( h \right) + 20 \ \log_{10} \left( \frac{ 1 }{ r } \right)
$$
where we have used 1 m as a normalizing length defining at the source. In the following, we examine the decrease in intensity levels with distance at three different altitudes, 10 km, 50 km, and 100 km.
If we move ~3 m away from the source, the intensity levels drop to ~85 dB, ~65 dB, and ~39 dB for, respectively. At ~10 m away, these intensities drop to ~74 dB, ~54 dB, and ~28 dB, respectively. At ~50 m away, these intensities drop to ~60 dB, ~40 dB, and ~14 dB, respectively. And at ~150 m away, the intensities drop to ~51 dB, ~31 dB, and ~5 dB, respectively. For comparison at sea level, the intensities would be ~90 dB, ~66 dB, and ~56 dB at distances of ~3 m, ~50 m, and ~150 m, respectively.
Thus, at 100 km altitude one need only move a little over 100 m away from the source before the intensity level drops below the hearing threshold (i.e., ~5 dB for a 20 year old male at 1000 Hz).
Answer 2
The model only went to 100 km but even so, our source would become difficult to hear if we moved a little more than ~100 m from it. Given that the density decreases exponentially with an e-folding distance of only ~8.5 km (pressure does so similarly as well), if we extrapolate our estimates for $L_{i,src}\left( h \right)$ then the value drops to ~10 dB by ~177 km.
So by ~200 km a human probably could not hear a source ~1 m away that produced a 100 dB, 1000 Hz intensity level at sea level.
From the Wikipedia article on sound:
In physics, sound is a vibration that propagates as a typically
audible mechanical wave of pressure and displacement.
To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave,
$$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial t^2}$$
(here $c$ is the speed of sound) but also the variation in the flow of air, that is, a particle displacement wave:
We consider the air in the tube to have a rest position, and the wave motion is expressed in terms of displacement from that position. If we denote by $ξ(x,t)$ the displacement of the air at position x at time t, then the wave equation for displacement is (I added a derivation at the end of this answer):
$$\frac{\partial^2 ξ}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 ξ}{\partial t^2}$$
As you cited, the pressure at the open end must be equal to the pressure of ambient air outside the pipe: this is just a boundary condition (I'll come back to this later, but a simple way to understand it is: if we didn't satisfy continuity of pressure, we'd have an infinite pressure gradient and so infinite force). However, for an open end, air can flow freely in and out: that is, the displacement wave does exit and propagate. This diagram shows what happens in an open (left) vs closed pipe (right) of the same length:
The red line is the amplitude of the variation in pressure, whereas the blue line is the amplitude of the variation in the flow of air. As you can see, the amplitude of the displacement wave is a maximum at the open end.
Now, to answer your questions, "why the pressure at the open end cannot vary from the atmospheric pressure?" and "how does the reflection of the wave at the open end occur?", I'll cite a passage from this wonderful explanation I've found (you should take a look at the animations if this is not clear enough, or alternatively at this video): Let's send a pulse of air down a cylindrical pipe open at the end. Then...
It reaches the end of the tube and its momentum carries it out into
the open air, where it spreads out in all directions. Now, because it
spreads out in all directions its pressure falls very quickly to
nearly atmospheric pressure (the air outside is at atmospheric
pressure). However, it still has the momentum to travel away from the
end of the pipe. Consequently, it creates a little suction: the air
following behind it in the tube is sucked out (a little like the air
that is sucked behind a speeding truck).
Now a suction at the end of the tube draws air from further up the
tube, and that in turn draws air from further up the tube and so on.
So the result is that a pulse of high pressure air travelling down the
tube is reflected as a pulse of low pressure air travelling up the
tube. We say that the pressure wave has been reflected at the open
end, with a change in phase of 180°.
For a more formal treatment of reflections at an open end, see this. Let me know if I can make myself more clear on any point.
Addendum: A derivation of the two wave equations.
The acoustic pressure is the local pressure deviation from the ambient, that is:
$$p(x,y)=P(x,t)- p_a$$
where $p_a$ is the ambient air pressure, and $P(x,t)$ is the absolute pressure. In this situation, Hooke's law states that
$$p=-B \frac {\partial ξ}{\partial x}$$
where the constant $B$ is the bulk modulus of air (i. e., air's resistance to uniform compression). Newton’s second law of motion implies that
$$\frac {\partial p}{\partial x} = - p_a \frac{\partial^2 ξ}{\partial t^2}$$
Combining these equations, we get
$$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial t^2}$$
and
$$\frac{\partial^2 ξ}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 ξ}{\partial t^2}$$
where $c = \sqrt{B/{\rho}}$ (from Newton-Laplace equation), $\rho$ being air density. Source of this derivation: Dave Benson's Music: A Mathematical Offering.
Later addendum:
As @knzhou pointed out in his answer, it should be mentioned that for an open end, $p = 0$ is really only an approximation, because the volume of air just outside of the tube is not infinite. However, you can make a more accurate representation of the actual tube by working in terms of an effective length, which is a little longer than the geometrical end. This diagram shows the effective length for the fundamental vibrational mode of a flute:
The end correction is the amount by which the effective length exceeds the actual length, and under normal conditions it is usually somewhere around three fifths of the width of the tube.
Also, one could wonder what is actually happening at the boundary between the tube and the open air. All I know is that if inside the tube there is a plane wave, and when the displacement wave exits and propagates externally, it becomes a spherical one, then there must be some complicated effect going on between the two (as was suggested by @docscience in the comments). Perhaps a more knowledgeable person can explain it to us.
Best Answer
I'd like to start from mathematical description of the following process. At the end of my post I will give you a qualitative explanation.
Mathermatical description of the motion in frequently oscillating field
Actually, the problem you state can be generalized to the following one:
Suppose you have a particle that moves in some force field that oscillates frequently in time. One can write the force, acting on the particle as follows: $$\mathbf{F}(\mathbf{r},t)=\mathbf{f(\mathbf{r})}\cdot \cos(\omega t)$$ Note, that the amplitude of oscillations depends on the position - in is a key feature. The question is to find out what the motion of that particle looks like, if the frequency of oscillations is very large.
The theory of motion in such a field is a well-known (and solved) problem. The detailed solution can be found in the following book: Landau, Lifshitz: Course of Theoretical Physics, Volume I "Mechanics" in paragraph 30. I won't retell you the calculations from the book in detail, but I will say several words about the result obtained there. Using some tricky math authors find "averaged" equations of motion for a particle in this case. This equations do not contain rapidly oscillating terms.
To be more exact, it is shown that instead of oscillating field one can introduce constant (in time) field with effective potential: $$U_{eff}=\frac{\mathbf{f}^2(\mathbf{r})}{4m\omega}$$ Here $\mathbf{f(\mathbf{r})}$ is amplitude of oscillating force. As you can see, if $\mathbf{f(\mathbf{r})}=const$ this effective potential is constant. Thus, there is no force in average, that acts on the particle. If the amplitude is not constant, then potential of this force is minimal there were $\mathbf{f(\mathbf{r})}^2$ is minimal. Thus, this forces pushes the particle to the place, where the amplitude is low.
In your case this force field is associated with pressure. Your problem can be treated in one dimension: there is pressure $p(x,t)=p_0\sin(kx)\cos(\omega t)$ that results in force $F(x,t)=-V\frac{\partial p(x,t)}{\partial x}=-Vkp_0\cos(kx)\cos(\omega t)$ (this is Archimedes law), where $V$ is the volume of a droplet. The effective potential, induced by the pressure field is: $$U_{eff}=\frac{V^2k^2p_0^2\cos ^2(kx)}{4m\omega}$$ So, the total potential energy (including gravitational one) is: $$U=-mgx+\frac{V^2k^2p_0^2\cos ^2(kx)}{4m\omega}$$ If the pressure amplitude is large enough, there are points of local minimums - equilibrium positions for droplets! (on the picture you can see illustration of this fact; it is potential enery against distance in some units; the average negative slope is associated with gravity and minimums are associated with pressure)
Qualitative explanation
But yet: how can one understand this result qualitaively? The explanation is as follows:
Suppose your particle moves around in vibrating pressure field in some way. It is quite clear that one can express the position of the particle (as a function of time) as follows: $x(t)=X(t)+\delta x(t)$, where $X(t)$ is smooth function and $\delta x(t)=-A\cos(\omega t)$ is small deviation that happens in antiphase to the field's oscillations (as always for excited oscillations with no friction). Imagine that your particle is now in the region of non-constant (in space) oscillating force and amplitude of this oscillations is higher in front of the particle and lower behind it. As was mentioned, $\delta x(t)$ happens in antiphase to external field, thus, when $\delta x>0$ the particle is pushed back and when $\delta x<0$ the particle is pushed in front. However, external field is higher in the region where $\delta x>0$ and lower where $\delta x<0$, so the total impulse over the period will be directed in the region of "weak" oscillations.
That's it. If you want, I can explain something in more detail.
PS: Sorry for my language, if it was a problem.