Disproportionation is a redox reaction. Since
$$\Delta G = -zF\Delta E$$
you may compare the redox potentials for both half-reactions in order to determine whether disproportionation occurs spontaneously ($\Delta G < 0$).
Note that nitrogen dioxide ($\ce{NO2}$) exists in equilibrium with dinitrogen tetroxide ($\ce{N2O4}$)
$$\ce{2 NO2 <=> N2O4}$$
and literature values for redox potentials are often given for $\ce{N2O4}$ (values at $\mathrm{pH}=14$):
$$\begin{alignat}{2}
\ce{N2O4 + 2 e- \;&<=> 2 NO2- }\quad &&E_1^\circ = +0.867\ \mathrm{V}\\
\ce{2 NO3- + 2 H2O + 2 e- \;&<=> N2O4 + 4 OH-}\quad &&E_2^\circ = -0.86\ \mathrm{V}
\end{alignat}$$
The potential $E_1^\circ = +0.867\ \mathrm{V}$ for the reduction of $\ce{N2O4}$ is larger than the potential $E_2^\circ = -0.86\ \mathrm{V}$ for the oxidation of $\ce{N2O4}$; therefore, disproportionation should occur.
In equlibrium:
$\begin{align}
E_1^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},1}}{a_{\text{red},1}} &= E_2^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}}\\
E_1^\circ - E_2^\circ &= \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}} - \frac{0.059\ \mathrm{V}}{z}\log \frac{a_{\text{ox},1}}{a_{\text{red},1}} \\
\left(E_1^\circ - E_2^\circ\right) \cdot \frac{z}{0.059\ \mathrm{V}} &= \log \frac{a_{\text{ox},2} \cdot a_{\text{red},1}}{a_{\text{red},2} \cdot a_{\text{ox},1}} \\
\left(0.867\ \mathrm{V} + 0.86\ \mathrm{V}\right) \cdot \frac{2}{0.059\ \mathrm{V}} &= \log\frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{\left[\ce{N2O4}\right] \cdot \left[\ce{N2O4}\right]} \\
{10^{58.5}} &= \frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{{\left[\ce{N2O4}\right]}^2} \\
{10^{29}} &= \frac{\left[\ce{NO3-}\right] \cdot \left[\ce{NO2-}\right]}{{\left[\ce{N2O4}\right]}}
\end{align} $
Therefore, $\ce{N2O4}$ is unstable with respect to disproportionation:
$$\ce{N2O4 + 2 OH- -> NO2- + NO3- + H2O}$$
