I have just realised that the simplest solution is to use an $(n-2)$-gonal bipyramid. That is to say, take two pyramids that have a regular $(n-2)$-gon as their base, and stick them together base to base. The vertices around the shared base all have degree $4$, so there are at most two vertices with odd degree (the pyramid tips), and it will have a Eulerian path or circuit. It is a convex polyhedron, so its edge graph is planar. It has $3(n-2)$ edges, so its edge graph is maximal.
Previous solution follows below.
Here I'll describe a construction, based on ArsenBerk's answer, which works for any number of vertices $n>4$. It constructs a planar graph with triangular faces, and with an Eulerian path or circuit.
Start with an $n$-gon. We now triangulate it twice, but the way we do it is slightly different for even and for odd $n$. For even $n$ we use two "zig-zag" triangulations as illustrated here for $n=8$.
The two triangulations start and end at the same pair of opposite vertices, so every vertex has even degree, and the graph has an Eulerian circuit. This easily generalises for all $n>4$, but fails for $n=4$ since in that case the two triangulations degenerate to be identical, making the graph not simple.
For odd $n$ you can use a fan-like triangulation, adding all diagonals from one of the vertices. The two triangulations must fan out from adjacent vertices because otherwise you get a duplicate edge. Here is an illustration for $n=7$.
The two adjacent vertices have degree $2+n-3=n-1$, which is even. The two vertices next to them have degree 3, and all the others have degree 4. The graph has only two odd degree vertices, so has an Eulerian path.
These graphs are planar. You can draw them so that one triangulation is inside the $n$-gon, and the other inverted on the outside. You can kind of think of it as a degenerate convex polyhedron, consisting of an $n$-gon around its equator with one triangulation on top and another on the bottom. As with a normal convex polyhedron you can expand a face in the top half to get a planar graph of the polyhedron edges.
Every face is triangular, so we cannot add any further edges to the graph. In fact, no graph is possible with more edges, and this can easily be proved using Euler's formula. In a planar graph every face has at least $3$ edges, and every edges borders $2$ faces, so $3F\le2E$. Putting this into Euler's formula $V-E+F=2$ you get $E\le3V-6$. In these graphs the number of edges is $n+2(n-3) = 3n-6$, so this maximum is attained.