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.
To expand on Xcheckr's answer:
The full equation for a single-frequency traveling wave is
$$f(x,t) = A \sin(2\pi ft - \frac{2\pi}{\lambda}x).$$
where $f$ is the frequency, $t$ is time, $\lambda$ is the wavelength, $A$ is the amplitude, and $x$ is position. This is often written as
$$f(x,t) = A \sin(\omega t - kx)$$
with $\omega = 2\pi f$ and $k = \frac{2\pi}{\lambda}$. If you look at a single point in space (hold $x$ constant), you see that the signal oscillates up and down in time. If you freeze time, (hold $t$ constant), you see the signal oscillates up and down as you move along it in space. If you pick a point on the wave and follow it as time goes forward (hold $f$ constant and let $t$ increase), you have to move in the positive $x$ direction to keep up with the point on the wave.
This only describes a wave of a single frequency. In general, anything of the form
$$f(x,t) = w(\omega t - kx),$$
where $w$ is any function, describes a traveling wave.
Sinusoids turn up very often because the vibrating sources of the disturbances that give rise to sound waves are often well-described by
$$\frac{\partial^2 s}{\partial t^2} = -a^2 s.$$
In this case, $s$ is the distance from some equilibrium position and $a$ is some constant. This describes the motion of a mass on a spring, which is a good model for guitar strings, speaker cones, drum membranes, saxophone reeds, vocal cords, and on and on. The general solution to that equation is
$$s(t) = A\cos(a t) + B\sin(a t).$$
In this equation, one can see that $a$ is the frequency $\omega$ in the traveling wave equations by setting $x$ to a constant value (since the source isn't moving (unless you want to consider Doppler effects)).
For objects more complicated than a mass on a spring, there are multiple $a$ values, so that object can vibrate at multiple frequencies at the same time (think harmonics on a guitar). Figuring out the contributions of each of these frequencies is the purpose of a Fourier transform.
Best Answer
Yes. The main reason sound decreases in amplitude with distance is not due to absorption; it's because sound sources emit roughly spherical radiation and are subject to the inverse square law. So an ideal plane wave in an ideal atmosphere would not attenuate with distance, because it's not subject to inverse square law.
But sound is also absorbed by the air, so even a plane wave will slowly decrease in amplitude with distance. This means the sound energy is being turned into heat energy, and increasing the temperature of the air slightly. The amount of energy absorbed varies with humidity and affects high frequencies first:
Absorption of Sound in Air versus Humidity and Temperature
Damping of Air of High Frequencies
Damping of Air of High Frequencies (Dissipation)