Yes, the particles are just moving back and forth, not moving in any particular direction on average. Here's an animation to make this more clear: http://www.acoustics.org/press/151st/Lindwall.html (first one on the page)
However, there are at least two weird, unphysical things about that animation. For one, the particles are only moving in response to the sound wave, in perfect synchronization, and they aren't also moving constantly in random directions. If you had some material that would remain a gas at absolute zero, then a sound wave in that absolute-zero gas would look more or less like the animation. But for an everyday-volume sound wave in room-temperature air, the random thermal motion (which is always happening whether or not there is a sound wave) is much stronger than the motion caused by the sound wave. The only reason we notice a sound wave at all is because it is an ordered motion that carries energy in a particular direction. If you followed the motion of a single air molecule, it would look entirely random and there would be no trace of the sound wave. The sound wave only becomes apparent when you look at the large-scale pattern of density variations.
The other weird thing about the animation is that the molecules stop moving and turn around without colliding with anything! It should be obvious that in a real gas made of electrically neutral particles, there is no long-distance force that would cause this to happen, and a moving particle would not change direction unless it actually collided with another particle.
To answer your particular questions, yes, the whole medium is "vibrating" in that the density and pressure are increasing and decreasing periodically, as the gas flows back and forth. However, I would not refer to it as a "tunnel", because, as you can see from the animation, there doesn't have to be any well-defined cross-sectional shape for the sound wave. In fact, the simplest geometry to consider is a plane wave, which extends infinitely in all directions perpendicular to the direction of propagation. It's actually impossible to make a "beam" of sound that will propagate forever without spreading out.
Do low frequencies carry farther than high frequencies? Yes. The reason has to do with what's stopping the sound. If it weren't for attenuation (absorption) sound would follow an inverse square law.
Remember, sound is a pressure wave vibration of molecules. Whenever you give molecules a "push" you're going to lose some energy to heat. Because of this, sound is lost to heating of the medium it is propagating through. The attenuation of sound waves is frequency dependent in most materials. See Wikipedia for the technical details and formulas of acoustic attenuation.
Here is a graph of the attenuation of sound at difference frequencies (accounting for atmospheric pressure and humidity):
![sound absorption at various frequencies](https://i.stack.imgur.com/aLJBb.png)
As you can see, low frequencies are not absorbed as well. This means low frequencies will travel farther. That graph comes from this extremely detailed article on outdoor sound propagation.
Another effect that affects sound propagation, especially through walls, headphones, and other relative hard surfaces is reflection. Reflection is also frequency dependent. High frequencies are better reflected whereas low frequencies are able to pass through the barrier:
![sound frequency transmission versus reflection](https://i.stack.imgur.com/sUghg.jpg)
This is and frequency-based attenuation are why low-frequency sounds are much easier to hear through walls than high frequency ones.
Frequency Loudness in Headphones:
The above description apply to sounds that travel either through long distances or are otherwise highly attenuated. Headphones start off at such low intensities already they don't travel long enough distances for attenuation to be a dominate factor. Instead, the frequency response curve of the human ear plays a big role in perceived loudness.
The curves that show human hearing frequency response are called Fletcher–Munson curves:
![Fletcher-Munson_curves](https://i.stack.imgur.com/6JUxQ.png)
The red lines are the modern ISO 226:2003 data. All the sound along a curve is of "equal loudness" but as you can see, low frequencies must be much more intense to sound equally as loud as higher frequency sounds. Even if the low frequencies are reaching your ear, it's harder for you to hear them.
Headphone sound is doubly compounded by the difficulty of making headphones with good low-frequency response. With loudspeakers you can split the job of producing frequencies among a subwoofer, a midrange speaker, and a tweeter. For low frequencies subwoofers are large and have a resonating chamber which simply isn't an option with headphones that must produce a large range of sound frequencies in a small space. Even a good pair of headphones like Sennheiser HD-650 struggle with lower frequencies:
![headphones frequency response](https://i.stack.imgur.com/GAoJs.jpg)
So if it sounds like high frequencies travel farther with headphones, it's because headphones are poor at producing low frequencies and your ear is poor at picking them up.
Best Answer
Sound pressure level (SPL) in dB is defined relative to a pressure $p_{ref}=20\mu Pa$ $$L_p=20\log_{10}\left(\frac{p_{rms}}{p_{ref}}\right )$$ 75dB corresponds to acoustic pressure of 0.11 Pa, you can use this online calculator to easily check other SPLs.
Acoustic velocity is proportional to acoustic pressure through acoustic impedance $Z=\rho c$ where $\rho$ is air density and $c$ is sound velocity, for air at room temperature $Z \approx 400 \frac{Ns}{m^3}$.
With acoustic pressure and impedance you can calculate all sorts of quantities. In particular, particle displacement is calculated as $$ \xi =\frac{p}{Z\omega}=\frac{p}{Z2\pi f} $$ where $f$ is the acoustic frequency. Plugging all numbers into this formula, we get $\xi=4.4 \cdot 10^{-7}\,m$. Keep in mind that SPL is given in logarithmic scale, so if you take $L_p=150\,dB$ then $\xi= 0.0025\,m=2.5\,mm$.