Merge the days Sunday & Wednesday call this node Suwday. Now Suwday, Monday, Tuesday $\color{blue}{3}$ and Thursday, Friday, Saturday $\color{red}{3}$ will need to form a $K_{\color{blue}{3},\color{red}{3}}$.
And thus by Kuratowski's Theorem https://en.wikipedia.org/wiki/Kuratowski%27s_theorem the original graph cannot be planar.
Taking the dual would give us a $4$-regular planar graph where we want as many of the faces as possible to be triangles. There's a family of $4$-regular polyhedra where almost all faces are triangles: the antiprisms. So the dual of an antiprism (apparently, this is called a trapezohedron) should be a good candidate for reaching the upper bound here.
For any even $n$, the dual of the antiprism consists of an $(n-2)$-cycle with two more vertices: one adjacent to every even vertex of the cycle, and one adjacent to every odd vertex. This has two vertices of degree $\frac n2-1$, and $n-2$ vertices of degree $3$. (When $n=8$, all $n$ vertices have degree $3$, and this graph is just the cube.)
Except when $n=8$, we cannot have $n$ vertices of degree $3$, but it's conceivable that we can have $n-1$ vertices of degree $3$ and one vertex of degree $n-5$. Let's analyze this case.
Let $v$ be the vertex of degree $n-5$. We know the bipartition, up to two cases: if we put $v$ on one side, and its $n-5$ neighbors on the other, then either the remaining $4$ vertices are all on the same side as $v$, or one of them is on the side with $v$'s neighbors and adjacent to the other $3$. In the first case, counting the edges from each side, we want $(n-5) + 4\cdot 3 = (n-5) \cdot 3$, so $n=11$. This is impossible by computer search, and there should be a combinatorial proof, but I haven't found a clean argument. In the second case, we want $(n-5) + 3 \cdot 3 = (n-4) \cdot 3$, so $n=8$, and that's the cube again.
One remaining question: can we have $n-2$ vertices of degree $3$ when $n$ is odd? Experimentally, no: here's a table of the maximum value of $n_3$ for small odd $n$.
\begin{array}{c|ccccccc}
n & 11 & 13 & 15 & 17 & 19 & 21 \\
\hline
n_3 & 8 & 10 & \color{red}{11} & 14 & 16 & \color{red}{17}
\end{array}
For $n=15$ and $n=21$, we can't even get to $n-3$ vertices of degree $3$.
There's a reason to expect a repeating pattern mod $6$: if you can get a solution for $n$ where two vertices $v,w$ of degree greater than $3$ share a face but are not adjacent, you can get a solution for $n+6$ by adding a cube graph on $v$, $w$, and $6$ new vertices, and this can be iterated. It's hard to say what's up with the upper bounds, though.
Best Answer
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.)