= Solution
Put $K=\sum_aN_aN_{\bar a}$. The <Verlinde formula> diagonalizes every <anyon fusion matrix> using the same <modular S-matrix>:
$$
N_a=S\,\operatorname{diag}_x\left(\frac{S_{ax}}{S_{1x}}\right)S^\dagger.
$$
Because $S_{\bar a x}=S_{ax}^*$, the <eigenvalue> of $K$ in column $x$ is
$$
\kappa_x=\sum_a\frac{S_{ax}S_{\bar a x}}{S_{1x}^2}=\frac{\sum_a|S_{ax}|^2}{S_{1x}^2}=\frac1{S_{1x}^2}.
$$
The last equality uses <unitary matrix> orthogonality. Therefore
$$
(K^g)_{11}=\sum_x|S_{1x}|^2\kappa_x^g=\sum_xS_{1x}^{2-2g}.
$$
Finally, insert $S_{1x}=d_x/\mathcal D$, where $d_x$ is the <quantum dimension> and $\mathcal D^2=\sum_xd_x^2$ is the squared <total quantum dimension>:
$$
\boxed{\operatorname{GSD}(\Sigma_g)=\mathcal D^{2g-2}\sum_xd_x^{2-2g}=\left(\sum_xd_x^2\right)^{g-1}\sum_xd_x^{2-2g}.}
$$
Although only <quantum dimensions> remain, the result is an integer because <2/a> counts an <anyon fusion space>.
The <sphere> value is $\mathcal D^{-2}\sum_xd_x^2=1$, and the <torus> value is $\sum_x1$. If all $n$ simple <anyons> have <quantum dimension> one, the answer is $n^g$, agreeing with the Abelian handle count. These checks also fix the exponent $2g-2$.
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