The real reason that you don't count rotation around the molecule's narrow axis is that that degree of freedom is frozen out. The formula for rotational kinetic energy that is relevant here is given by:
$$K = \frac{1}{2} \vec{L} \cdot I^{-1} \vec{L},$$
where $I$ is the moment of inertia tensor. If we choose our coordinates to diagonalize $I$ for a molecule we get (ignoring numerical factors fixed by the geometry):
$$I \sim m\left[\begin{array}{ccc}
a_0^2 & 0 & 0 \\
0 & a_0^2 & 0 \\
0 & 0 & r_{\mathrm{nuc}}^2
\end{array}\right],$$
where $a_0$ is the Bhor radius ($\approx 5\times10^{-11}\operatorname{m}$), $r_{\mathrm{nuc}}$ radius of the nucleus ($\approx 1\times 10^{-15}m$), and $m$ is the mass of the atoms (basically the nuclei).
Because angular momentum is quantized in packets of size $\hbar \approx 10^{-34} \operatorname{m}^2\operatorname{kg} \operatorname{s}^{-1}$ it takes about $K \sim \frac{\hbar^2}{m a_0^2} \approx 7\operatorname{meV}$ (that's milli-electron-Volts, $T\sim 90 \operatorname{K}$) to set an $\mathrm{H}_2$ molecule spinning about its long axes, but about $K \sim \frac{\hbar^2}{m r_{\mathrm{nuc}}^2} \approx 7\operatorname{MeV}$ of energy (mega-electron-Volts, $T\sim 8 \times 10^{10}\operatorname{K}$) to set it spinning about its narrow axis. Needless to say, that much thermal energy would quickly dissociate the molecule and ionize the atoms.
Bottom line, it takes about $10^{10}$ times as much energy to get a diatomic molecule spinning along its narrow axis than the long axes, making that degree of freedom very effectively frozen out.
It is correct that the equipartition theorem would, classically, put the lie to the claim that the energy stored in rotations around the narrow axis contain negligible energy - classically it would have its $kT/2$, too. This is one of those times that quantum mechanics is unavoidable for explaining a phenomenon.
