Consider the class of planar triangulations (allowing multi-edges). These are maximal planar graphs (for a fixed number of vertices), i.e. any planar graph can be completed to one. Then the maximal number of 4-colorings of a planar triangulation with $n$ vertices is $3\cdot 2^n$.
To see this, first assume that the triangulation no separating triangle. Whitney proved that any planar triangulation with no separating triangle has a Hamiltonian cycle. On either side of the Hamiltonian cycle, one sees a disk with an $n$-cycle on its boundary, and $n-3$ edges in the interior. The number of colorings of such a graph is $3\cdot 2^n$, since you can color a triangle with $4\cdot 3\cdot 2$ colors, and then the remaining $n-3$ vertices each have $2$ colors available working outwards. So this gives an upper bound on the number of colorings. To achieve this upper bound, take two copies of the triangulated disk, and glue them along the boundary by the identity. Then this is a planar triangulation with the same number of $4$-colorings (of course, there will be many multi-edges, so it is far from simplicial). This class of graphs is a natural class to consider, e.g. in the proof of the $4$-color theorem.
If there is a separating triangle, then one may show that the graph has $<3\cdot 2^n$ $4$-colorings by induction. If a triangulation has a separating triangle, then one may break it up into two triangulations, so that the number of $4$-colorings is the product of the number of colorings of each triangulation $/24$, and by induction one sees that the number of 4-colorings is strictly $<3 \cdot 2^n$.
Maybe you'll object to this choice of graphs, e.g. because of multi-edges. If you make no restriction, then obviously the trivial graph on $n$ vertices will be maximal. For connected graphs, trees will be maximal. Other natural classes are trivalent graphs, bipartite graphs, or simplicial triangulations. I'm not sure about the maximal number of colorings for these classes of graphs.
Addendum: Restricting to $3$-connected simplicial triangulations, the growth is still $O(2^n)$. Consider an $n$-cycle, then the number of $3$-colorings is $O(2^n)$. Then cone it off on both sides to get a simplicial triangulation with $O(2^n)$ $4$-colorings. I'm not sure what the optimal constant is though.