Summary: The comments are correct; dispersion is technically present but nearly imperceptible. However, strictly speaking the group velocity does increase with frequency.
Detailed Discussion:
According to Blackstock's book Fundamentals of Physical Acoustics, the wavenumber in a viscous fluid (thermal effects are accounted for in a nearly identical manner for constant frequency signals, and are of the same order as viscous effects) may be written as
$$ k = \frac{\omega/c_0}{\sqrt{1+i\tilde V\delta_v\omega}}, $$
where $k$ is the wavenumber, $\omega$ is the angular frequency, $c_0$ is the inviscid wave speed, $\tilde V=O(1)$ is the viscosity number, and $\delta_v=\nu/c_0^2$, with $\nu$ being the kinematic viscosity. For standard conditions $\delta_{v_{air}}\approx 1.3\times10^{-10}$ s. Taking the derivative of $k$ with respect to $\omega$ yields the inverse of the group velocity, $c_g$, and so we may write
$$c_g = \left(\frac{\partial k}{\partial\omega}\right)^{-1} = c_0\frac{2(1+i\tilde V\delta_v\omega)^{3/2}}{2+i\tilde V\delta v\omega}.$$
The imaginary part is associated with losses, so we want to focus on the real part. Noting that $\delta_v\omega$ is going to be very small compared to 1 or 2 for any signal of consideration, we may expand this function and write
$$\Re\left\{\frac{c_g}{c_0}\right\} = 1 + \frac{\tilde V^2\delta_v^2\omega^2}{8} + O(\delta_v^4\omega^4).$$
Thus, the group velocity does indeed become larger with frequency. However, the increase is negligible for all frequencies of interest. For example, ultrasound can go as high as the megahertz, so lets go even higher and use $\omega=10^{7}$ s$^{-1}$. Then we may write the increase of the group velocity from the zero-frequency limit as being on the order of $(10^{-10}\cdot10^{7})^2c_0=10^{-6}c_0$. This is on the order of millimeters per second compared to hundreds of meters per second. Slight fluctuations of the ambient temperature have a far more profound effect than the dispersion of the air.
