Rather than subdivide edges, another way to get bigger graphs while keeping the difference $|E|-|V|$ the same, and not affecting $k$-planarity, is to add leaf vertices. Most simply, just take a graph with $n$ vertices and $m$ edges, and add a path with $t$ new vertices and $t$ new edges from one of its vertices; the result has $n+t$ vertices and $m+t$ edges, and the new graph is $k$-planar if and only if the original graph was.
I do not think we will find infinitely many examples of non-$1$-planar graphs with $n$ vertices and $n+3$ edges, because I do not think we will find any. The question is: what is the best difference between edge and vertex count we can achieve?
I found many examples of non-$1$-planar graphs with $n$ vertices and $n+11$ edges. According to Wikipedia, $K_{3,7}$ (with $10$ vertices and $21$ edges), $K_{4,5}$ (with $9$ vertices and $20$ edges), $K_{1,3,4}$ (with $8$ vertices and $19$ edges), and $K_{1,1,1,1,3}$ (with $7$ vertices and $18$ edges) are not $1$-planar. This ultimately comes from 1-planarity of complete multipartite graphs by Czap and Hudák. Also, Minimal non-$1$-planar graphs by Korzhik points out that $K_{1,1,1,1,3} = K_7 - K_3$ is a minimal non-$1$-planar graph, and that subdividing one of the edges between its degree-$6$ vertices gives another minimal non-$1$-planar graph (with $8$ vertices and $19$ edges).
Of course, we can use the path-adding technique to construct infinitely many non-$1$-planar graphs with $n$ vertices and $n+11$ edges. I'm pointing out that there's a bunch of $n+11$ examples because I think it's evidence that we can't reach $n+10$; if we hit the same wall from many different directions, we can suspect that the wall is everywhere. (But it's not very strong evidence.)
We can slightly generalize Euler's formula as: $V-E+F-C = 1$, where $C$ is the number of components. Most of the time $C=1$, which gives us the familiar formula. But it works great if $C>1$. And the question is about the "trivial" case where $C=0$. Now of course $V=E=0$ and $F=1$, giving us the right answer.
There is a standard inductive proof of Euler's formula which involves either removing an edge and a face, or removing an edge and a vertex, and observing that the invariant remains the same. See here. But these proofs are quite complicated, because they are trying to preserve connectedness.
With the generalized formula, the proof can be much cleaner (still skating over elementary topology). Remove any edge. There are two disjoint possibilities:
(1) We merge two distinct faces, so: $E-1; F-1$
(2) The edge always had the same face on both sides, so removing the edge instead splits a component into two: $E-1; C+1$
In either case the invariant $V-E+F-C$ remains the same. So we remove all the edges, and then have a bunch of isolated vertex/components in a single face, so $V=C$ and hence $V-0+1-C = 1$.
Always nice when a formula can apply all the way down to $0$.
Best Answer
Let $G$ be the planar graph in this drawing. As you've found, it has $n + \binom n4$ vertices. It has the original vertices of $K_n$; beyond that, for any four vertices $a,b,c,d$ of $K_n$, exactly one of the intersections $ab \cap cd$, $ac \cap bd$, or $ad \cap bc$ exists (and is a vertex of $G$).
We can also figure out the number of edges of $G$. Each of the first $n$ vertices has degree $n-1$; each of the other $\binom n4$ vertices has degree $4$. Add up the degrees, and you get twice the number of edges.
Finally, Euler's formula $v - e + f = 2$ will tell us the number of faces of $G$ once we know the number of vertices and edges.