What is the value of air anisotropy?

What you're asking about is the air adiabatic index (which I'll call $\gamma$) rather than anisotropy.

Not to get too hung up about the Greek, but adiabaticity is about how it a gas changes under transformations that don't include heat transfer, while anisotropy is about how asymmetric something is under rotations - something that doesn't seem to make much sense for a gas. Note also that isentropic (literally - constant entropy) is a synonym of adiabatic for all practical purposes. I don't blame anyone for getting confused about all this Greek.

Luckily, $\gamma$ is pretty easy to calculate, it's simply $\tfrac{C_p}{C_p−R}$ where $R=8.314$ J/(mol K) is the ideal gas constant. The heat capacity $C_p$ can easily be looked up for any molecule you might care about, chemists have been compiling that kind of data for a century or two. As air is a mixture, you have to calculate its $C_p$ by taking the average of the heat capacity of its components (weighted by molar ratio, not mass ratio). I realise it's also possible that OP meant "exhaust gases" rather than air specifically. In this case, the heat capacity can't be determined without knowing exactly what was being burnt, in what proportion, and with what combustion efficiency. Still, as water is one of the most common exhaust gases for many different kinds of fuel, I've included it in the graph below.

Unluckily, $C_p$ and $\gamma$ are functions of temperature. This isn't a huge dependency, but when going from combustion chambers at 1000s of Kelvins to ambient, it can matter. Still, we can compute $\gamma(T)$ easily enough, which I've done here (mathematica webbook for calculations and sources). The end result is:

Note that here "air" means dry air. If you have moisture content in the air, this will change; the calculations can be easily modified to take into account a known fraction of water content. I also didn't plot steam below 373K, because at that point (depending on pressure) it condenses to a liquid and phase transitions make everything far more complicated.

The final question that remains is: what value of $T$ the OP should use for the $\gamma$ in their equation? This is slightly problematic, because the derivation of that equation assumes an ideal gas with constant heat capacity. However, a reasonable heuristic is that we should use the "hot" $\gamma$ when looking at the energy each particle has in the combustion chamber, and the "cold" $\gamma$ when computing the efficiency of the nozzle after the exhaust has been expanded and cooled. Doing this, the relevant part of the equation in the OP becomes

$$ v_e = \sqrt{2 \frac{\gamma_c}{\gamma_c - 1} T_c \left(1 - \frac{P_e}{P_c}^\tfrac{\gamma_e - 1}{\gamma_e}\right)} $$

where $c$ is for the conditions at the combustion chamber and $e$ at the exhaust.

Or, ignore all that, and take the crude physicist calculation that $\gamma = \frac{5}{3}$ for anything monatomic, $\tfrac{7}{5}$ for diatomic gases ($\tfrac{4}{3}$ when "hot", $\tfrac{5}{4}$ when "very hot"); and anything more complex starts at $\tfrac{4}{3}$ and drops with increased temperature (with ~$1.1$ being the effective floor).