$H^*(Y_n;\mathbb{Q}) \cong (H^*((S^3)^N;\mathbb{Q}))^{\Sigma_n}$.
Hence with rational coefficients this reduces to analyzing the invariant subspace of the action of $\Sigma_n$ on $H^*((S^3)^N;\mathbb{Q})\cong (\mathbb{Q}^2)^{\otimes N}$.
I suspect that this will correspond to the number of simple graphs with $n$ vertices. https://oeis.org/A000088 A graph on the vertex set $\{1,\ldots,n\}$ gives a choice of cohomology class in $H^*((S^3)^N;\mathbb{Q})$, where for vertices $\{i, j\}$ one chooses a generator of $H^0(S^3)$ if $\{i,j\}$ does not span an edge, and the generator of $H^3(S^3)$ if it does, then mulitiply together via the Kunneth isomorphism $H^*((S^3)^N)\cong H^*(S^3)^{\otimes N}$. Hence one obtains an element of $H^*(Y_n)$ by averaging over the action of $\Sigma_n$. I suspect that these classes are linearly independent for different orbits of graphs, which would prove that the dimension of $H^*(Y_n)$ is the number of isomorphism classes of simple graphs on $n$ vertices, but I haven’t checked it.