If $n$ is odd then there are no intersections of $3$ or more diagonals and the problem can be analysed quite simply. Let's define a diagonal to be of rank $k$ ($k=0, \ldots, (n-1)/2$) if between its two endpoints $k$ vertices are comprised. Of course rank $0$ diagonals are the polygon sides, which form the outer circle. A diagonal of rank $1$ intersects all the $n-3$ diagonals issued from the single point it comprises, so we have $n-3$ intersection points. But these points are disposed symmetrically on the diagonal, so they have the same distance from the center in pairs and belong to only $(n-3)/2$ different circles.
Let's now consider a rank $2$ diagonal. We want to count its intersections with the other diagonals having rank $2$ or more, because its intersections with rank $1$ diagonals have already been counted before. That diagonal intersects all the $2(n-5)$ diagonals of rank $\ge2$ issued from the two point it comprises, but by symmetry we have $2(n-5)/2$ different new circles.
This analysis can be repeated, so for the total number $N$ of circles we have:
where the initial $1$ is due to the outer circle. That is:
However, if your results are correct, this formula fails for $n=15$ but I haven't explicitly checked that case.
If $n$ is even the analysis becomes much more complicated due to the presence of multiple diagonal intersections, see the paper cited by Lucian above.