Why are sound waves adiabatic?

For starters, to quote Allan Pierce in Acoustics,

The often stated explanation, that oscillations in a sound wave are too rapid to allow appreciable conduction of heat, is wrong.

That one surprised me when I learned it myself.

In fact, sound is not an adiabatic process for all frequencies. For any medium there is a thermal conduction frequency, $$ f_{\mathrm{TC}} =\frac{\rho c_{p} c^{2}}{2 \pi \kappa}. $$ Frequencies much lower than this value will be well-approximated as adiabatic. However, increasing the frequency through and above this point will transition the process from adiabatic to isothermal. For air, this frequency is $\sim 10^{9} \, \mathrm{Hz}$, well above the range of human hearing, so we almost always treat sound as adiabatic.

The physical reason this occurs is that heat transfer due to conduction is proportional to the temperature gradient. This is just a statement of Fourier's law for heat conduction. Consider what happens as the frequency of a harmonic wave decreases: The wavelength increases, and the slope of the oscillating waveform decreases as it is "stretched out." Assuming equal amplitudes, lower frequency waves will therefore set up smaller temperature gradients, which will conduct heat less effectively. If the heat conduction is negligible, then the entropy is conserved by the process.

So, in summary, the thermal gradients set up by sound waves for typical frequencies of interest are small enough to be neglected, hence sound is a (very nearly) adiabatic process. However, as Thomas pointed out below, in reality frequencies that cross into the potentially-isothermal regime are almost always affected by attenuation first, and the principal effects from conduction and viscosity are actually to damp out the sound wave.

In case you decide you do want to see some math, the energy equation is $$ \rho T \frac{d s}{d t} = \kappa \nabla^{2} T.$$ The previous arguments can be seen mathematically by linearizing about a quiescent base state and assuming harmonic wave solutions for $s$ and $T$. The equation can be rewritten as $$ - i \omega \rho_{0} T_{0} \hat{s} = - \kappa \frac{\omega^{2}}{c^{2}} \hat{T}, $$ $$ \hat{s} = - i \frac{\kappa \omega}{\rho_{0} T_{0} c^{2}} \hat{T}. $$ As the angular frequency $\omega = 2 \pi f \rightarrow 0,$ so must the amplitude of the entropy oscillation, $\hat{s}$. As with the quote at the beginning, much of my answer draws from Acoustics by Allan Pierce.

At the level of the Euler equation there is no heat flow between adjacent fluid elements (the thermal conductivity is zero). Because of $dQ=TdS$ this implies that the entropy in a co-moving fluid element is conserved (fluid elements can do $pdV$ work on one another). In equations $$ D(s/n) = \left( \partial_0+\vec{u}\cdot\vec\nabla\right) (s/n)=0, $$ where $s$ is the entropy density and $n$ is the particle density. Note that $s/n$ is the entropy per particle. Using the continuity equation this also implies $$ \partial_0 s+\vec\nabla(s \vec{u})=0 $$ which is usually called the equation of entropy conservation. This implies that we should relate the change in pressure $\delta P$ and the change in density $\delta\rho$ with $\rho=mn$ in a sound wave using the compressibility at constant entropy per particle $$ c_s^2 = \left.\frac{\partial P}{\partial\rho}\right|_{s/n=const}. $$ Now the only question is how this is modified when we consider the Navier-Stokes equation. This includes viscous friction and heat flow. Consider the relative size of time derivative $\partial_0 u$ and the viscous friction term $\eta/\rho\nabla^2 u$. For a plane wave $$ \frac{\partial_0 u}{\eta/\rho\nabla^2 u}\sim \frac{\eta}{\rho}\frac{k}{c_s} $$ which can be made arbitrarily small in the long wave length limit. The same is true for heat conduction. This means that the speed is approximately given by the adiabatic result, and that the approximation is arbitrarily good as $k\to 0$.