It depends on the mass of the molecule in question. Here's a quick, back-of-the-envelope answer. In a body at thermal equilibrium, every energy mode has the same average amount of energy, $\frac12kT$, where $T$ is temperature and $k$ is Boltzmann's constant. One of the energy modes is the translational kinetic energy of a molecule in some direction $x$, $\frac12mv_x^2$. We can solve
$$\frac12kT=\frac12mv_x^2$$
to find
$$v_x=\sqrt{\frac{kT}m}$$
and then plug in $k=1.38×10^{-23}\rm{m^2 kg s^{-2} K^{-1}}$, $T=300\rm{K}$, and for $m_{\rm{N}_2}=2×14\rm{u}=2×14×1.66×10^{−27} \rm{kg}=4.65×10^{−26} \rm{kg}$ to get
$$v_x=298\rm{m/s}=667mph.$$
The molecule is also moving in the $y$ and $z$ axes, so the answer depends on what exactly you mean by average speed: mean spead vs. root-mean-square speed.
This ignores rotational and vibrational degrees of freedom. Similar calculations may be performed for other substances.
Some links:
http://en.wikipedia.org/wiki/Root-mean-square_speed