As it stands there are 6 linearly independent terms to deal with ($a$, $a^2$, $h$, $h^2$, $h^3$ and $V$). Without being able to spot a common factor somewhere it's not clear one can accommodate so many terms (the three terms in $h$ in particular make it hard).
Saint-Robert's criterion indicates that it can't be done with three straight axes.
After playing around trying to simplify it (by trying to eliminate a term in various ways) in order to try to get it in a form dealt with by Clark, I think that perhaps Clark's approach may not be sufficient, which suggests one might have to go to Warmus' laborious criteria to even see if its possible as it stands, but I should go back and check it against either the Massau or Lecornu conditions first.
If you are allowed to reparameterize it to be in terms of say $p = \frac{h}{a}$ and $a$, or even easier, $q = 1 - \frac{h}{a}$ and $a$, then you can get somewhere.
If a good approximation is okay, it can be done with an order 4 (i.e. genus I) nomogram pretty closely, and one can generate better approximations by upping the order, possibly* to such a close approximation compared to the error in using an actual 'exact' nomogram that it wouldn't matter - typically anything below about half a percent error, though with care it can be a bit lower; if you get the worst approximation error much lower than that, you would have trouble telling an exact nomogram from an approximate one.
*(depending on what range of values you want to cover, I just assumed some values in order to play with it)
Edit: I see now that $\frac{V}{a^3}-\frac{1}{18}=(6\frac{h}{a}-1)^3/18$.
This suggests some possibility involving a pair of (possibly overlaid or back-to-back) $N$-charts but I don't currently see how to make it work without requiring multiple isopleth lines drawn - which is likely more effort for the end-user than it's worth.
Still, the observation may help someone spot a way to take it further.