Prediction of ionic reduction?
Fortunately, the explanation is not as complex as it might seem. It is an oft-made analysis, but one must be very careful when determining reaction spontaneity by looking at the standard free energies $\Delta G_r^o$ alone, as most reactions are not performed in standard conditions and therefore the actual reaction free energy change $\Delta G_r$ may be quite different.

You are correct that, in standard conditions, $\Delta G_r = \Delta G^o_r$ and therefore that $\ce{NO}$ is the most stable reduction product of $\ce{NO_3-}$. However, standard conditions entail, among other things, that $\ce{[H+]\ = \ 1\ M}$ (at least approximately). This corresponds to quite an acidic aqueous solution, with pH zero. If one were to examine the reaction in a more dilute acidic medium, then the value of $\Delta G_r$ for each reduction reaction would change. The reaction that produces $\ce{NO_2}$ has a $\Delta G_r$ that depends on $\ce{[H+]^2}$, while for $\ce{NO}$ the $\Delta G_r$ varies with $\ce{[H+]^4}$. It is clear then that, as one goes to a more and more dilute acidic medium, $\Delta G_r$ decreases more strongly for the reaction that produces $\ce{NO}$, and at sufficiently weak acidic solutions the relative stabilities of the products actually switches, so then $\ce{NO_2}$ becomes the most stable product. A very similar argument can be made for the switch between $\ce{H+}$ and $\ce{SO^{2-}_4}$ reduction in media with sulfuric acid. For $\ce{MnO^{-}_4}$, note that the conversion to $\ce{Mn^{2+}}$ implies the consumption of eight protons, and hence if it can happen, it will likely do so in acidic media. As the acidity of the solution decreases, the conversion to $\ce{Mn^{2+}}$ is quickly suppressed, favouring reactions for which the $\Delta G_r$ depends less strongly on $\ce{[H+]}$, and so we find that $\ce{MnO^_2}$ becomes more favourable.