Here, we need to use the Sedov-Taylor solution (the section was originally more detailed, but I see, regrettably, that edits have been made). You can fine the original paper by Taylor here, although it's not the best introduction to the solution; I'd recommend these notes instead.
The radius of the wave at a time $t$ is
for energy $E$, density $\rho$ and constant factor $\xi_0$. Differentiating, we find
It's safe to assume that $\xi_0\approx1$, for our purposes. Air density at sea level is roughly 1.225 kg/m$^3$. Now, the Trinity test released about 10 kilotons of TNT in the form of the blast. Plugging this in, we see that at $r=2$ m, we get $v=882,000$ m/s, while at $r=5$ m, we get $v=209,000$ m/s.
This is a lot higher than the value quoted in Mazura's answer, but I'll note that once we get out to about a quarter mile from the center of the explosion, the blast wave slows to speeds more like that one - in the hundreds of meters per second.
This is a wee bit simplified. If you want to look at the Sedov-Taylor solution in much more detail, I'd recommend scrolling through these slides, which are extremely detailed - and also not light bedtime reading.