I am just expanding abx's comment: it is possible to realize this branched cover as a group quotient. If abx prefers to explain this, I am happy to delete this answer.
Denote by $\mu_2$ the $2$-element group with an identity element $1$ and a non-identity element $-1$. There is an action of $\mu_2$ on $T^2$ by $-1\cdot (a,b) = (-a,-b)$. The quotient is $\mathbb{CP}^1$ considered as the projective space of nonzero, degree $1$, homogeneous polynomials in variables $x$ and $y$ up to scaling, $$q:T^2 \to \mathbb{CP}^1, \ \ q(a,b) = [r(a,b)x-s(a,b)y].$$$$q:T^2 \to \mathbb{CP}^1, \ \ q(a,b) = [r(a,b)y-s(a,b)x].$$ Now, for arbitrary $n$, let $\Gamma$ be the wreath product group $$\Gamma = \left( \mu_2 \right)^n \rtimes \mathfrak{S}_n, $$ where $\mathfrak{S}_n$ is the symmetric group on $n$ letters. The group For an element $G$$(g_1,\dots,g_n)\in (\mu_2)^n$ and for $\sigma\in \mathfrak{S}_n$, the element $\left((g_1,\dots,g_n),\sigma\right)\in \Gamma$ acts on $(T^2)^n$ in the evident manner, $$((g_1,\dots,g_n),\sigma)\cdot ((a_1,b_1),\dots,(a_n,b_n)) = \left((g_1\cdot (a_{\sigma(1)},b_{\sigma(1)}),\dots, g_n\cdot(a_{\sigma(n)},b_{\sigma(n)})\right).$$$$((g_1,\dots,g_n),\sigma)\cdot ((a_1,b_1),\dots,(a_n,b_n)) = \left(g_1\cdot (a_{\sigma(1)},b_{\sigma(1)}),\dots, g_n\cdot(a_{\sigma(n)},b_{\sigma(n)})\right).$$ The quotient of this group action is $\mathbb{CP}^n$ considered as the projective space of nonzero, degree $n$, homogeneous polynomials in variables $x$ and $y$ up to scaling , $$ q_n:T^{2n} \to \mathbb{CP}^n, \ \ q_n\left((a_1,b_1),\dots,(a_n,b_n)\right) =$$ $$ \left[(r(a_1,b_1)x-s(a_1,b_1)y)\cdots(r(a_n,b_n)x-s(a_n,b_n)y)\right]. $$$$ \left[(r(a_1,b_1)y-s(a_1,b_1)x)\cdots(r(a_n,b_n)y-s(a_n,b_n)x)\right]. $$