$H^*((S^3)^N/\Sigma_n;\mathbb{Q})$ is computed here.
It makes a little more sense to compute $H^*((S^2)^N/\Sigma_n;\mathbb{Q})$ given that global phase is irrelevant. The proof is exactly the same.
Still, that's not really the input space for quantum graphs. It makes more sense to use $N=\binom{n}{2}$ and $P(V^{\otimes N}) \cong \mathbb{CP}^{2^N-1}$ where $V \cong \mathbb{C}^2$.
Define an action of $\Sigma_n$ on $P(V^{\otimes N})$ as follows. Let $e_{12}, \ldots e_{n-1,n}$ be a basis for each $V$ such that $e_{ij}=e_{ij}^0|0\rangle+e_{ij}^1|1\rangle$ where $e_{ij}^k \in \mathbb{C}$. Then a basis for $V^{\otimes N}$ is given by $e_{12}^{k_{12}} \otimes \ldots \otimes e_{n-1,n}^{k_{n-1,n}}$ where $k_{ij} \in \{0,1\}$. $\sigma \cdot e_{12}^{k_{12}} \otimes \ldots \otimes e_{n-1,n}^{k_{n-1,n}} = e_{\sigma(1)\sigma(2)}^{k_{12}} \otimes \ldots \otimes e_{\sigma(n-1)\sigma(n)}^{k_{n-1,n}}$. We can act on $P(V^{\otimes N})$ by using projective coordinates $\sigma \cdot \left[ x_1, \ldots , x_{2^N}\right] = \left[ \sigma(x_1), \ldots, \sigma(x_{2^N})\right]$.
$H^*(\mathbb{CP}^{2^N-1}/\Sigma_n;\mathbb{Z}) = H^*(\mathbb{CP}^{2^N-1};\mathbb{Z})^{\Sigma_n}$. How do I compute this or adapt the proof linked above?
