I am not sure if this is the answer you are looking for, but a discussion on this topic can conclude upper and lower bounds for the number of edges in such a graph.
As you have already stated, we can determine the upper limit using the Euler characteristic, $ v-e+f-c=1 $ which we can reduce to $v-e+f=2$ since our graph is connected. Because we know that our original graph $G$ is non-planar before removing an edge, we can say that $\left|E\right|=e+1$. We can use the maximum possible faces of a planer graph with $v=15$ to find $e$ using $f_{max}=2v-4 = 26$
$$
15-e+26=2 \implies \left|E\right| \leq 40 \leftarrow\textrm{upper bound}
$$
To find the lower bound, we can use Kuratowski's Theorem (see here for a nice explanation) to state a lower bound, $\left|E\right| \geq v+3 = 18$.
So, we can say that for a non-planar graph $G$, with $\left|V\right|=15$ and the property that removing an edge must make the result planar, has
$
18\leq\left|E\right|\leq 40
$
edges.
Essentially, if the number of edges is less than the lower bound, we know that the graph must be planar. If the number of edges is greater than the upper bound, we would have to remove more than one edge to make it planar.