Henning once told me that,
[i]t follows from the Euler characteristic of the plane that the average face degree of a 3-regular planar graph with $F$ faces is $6-12/F$, which means that every 3-regular planar graph has at least one face with degree $5$ or lower.
I tried to understand and extend this and got the following:
Given a $k$-regular graph. Summing over the face degrees $f_n$ gives twice the number of edges $E$ and this is $k$ times the number of vertices $V$: $$ \sum f_n = 2E =kV \tag{1} $$ Further we have Euler's formula saying $$ V+F = E +\chi, $$ where $\chi$ is Euler's characteristic of the surface where the graph lives on. Again we insert $E=\frac k2V$ and get: $$ F=\left( \frac k2 -1 \right)V+\chi\\ V=\frac{F-\chi}{\frac k2 -1}. \tag{2} $$ Dividing $(1)$ by $F$ and inserting $(2)$ gives: $$ \frac{\sum f_n}{F}= \frac{k(F-\chi)}{(\frac k2 -1)F}=\frac {2k}{k -2} \left( 1-\frac{\chi}{F}\right) \tag{3} $$
or
$$ \sum f_n=\frac {2k}{k -2} \big( F-\chi\big). \tag{3$^\ast$} $$
Plug in $k=3$ and $\chi=2$ (the characteristic of the plane), we get back Henning's formula, but when e.g. $\chi=-2$, so the surface we draw could be a double torus, we get the average degree to be:
$$ 6 \left( 1+\frac{2}{F}\right) $$
How to find cubic graphs where the $\min(f_k)>6$ on surfaces with $\chi<0$?
EDIT The graph should not be a snark.
For orientable surfaces, here's a representative element of a family of non-snarky cubic graphs on an $n$-torus with $4n-2$ vertices, $6n-3$ edges and a single $(12n-6)$-sided face.
If it is a problem that some pairs of vertices have more than one edge going between them, that can easily be fixed with some local rearrangements (which can be chosen to preserve the green-blue Hamiltonian cycle and thus non-snarkiness).
For non-orientable surfaces, I think it is easiest to start by appealing to the classification theorem and construe the surface as a sphere with $k\ge 2$ cross-caps on it. Now if you start with a planar graph and place a cross-cap straddling one of its edges (so following the edge from one end to another will make you arrive in the opposite orientation), the net effect is to fuse the two faces the edge used to separate.
Therefore, start with an arbitrary planar non-snarky cubic graph with $2k-2$ vertices and $3k-3$ edges. Select an arbitrary spanning tree (comprising $2k-3$ edges), and place cross-caps on the $k$ edges not in the spanning tree. This will fuse all of the original graph's faces, and we're left with a graph with a single $(6k-6)$-sided face.
In each of the above cases, if you want more faces, simply subdivide the single one you've already got with new edges. The face itself, before you glue its edges together to form the final closed surface, is just a plain old $(12n-6)$- or $(6k-6)$-gon, so you can design the subdivision as a plane drawing.