Extension between the abelianization of the pure braid group and the symmetric group

The braid group $B_n$ and the pure braid group $P_n$ sits in a short exact sequence$$1\to P_n\to B_n\to S_n\to 1.$$The pure braid group $P_n$ has abelianization $\mathbb Z^{n\choose 2}$, with isomorphism (I believe) given by $P_n\to \mathbb Z^{n\choose2}$, where for any subset $\{i,j\}\subset\{1,\dots,n\}$, one obtains a map $P_n\to P_{\{i,j\}}$ which simply forgets all other strands, and pure braid group $P_{\{i,j\}}$ on two strands is isomorphic to $\mathbb Z$.

What is the quotient $B_n/[P_n,P_n]$?It sits in a short exact sequence$$1\to\mathbb Z^{n\choose 2}\to B_n/[P_n,P_n]\to S_n\to 1,$$which is classified by a class in $H^2(S_n,\mathbb Z^{n\choose2})$; what is it?

I believe it is not split (i.e., the cohomology class is nontrivial), since if the exact sequence split there would be a homomorphism $B_n\twoheadrightarrow \mathbb Z^{n\choose2}\rtimes S_n\to\mathbb Z$ which sends $(a_{ij})\in\mathbb Z^{n\choose2}$ to $\sum_{i<j}a_{ij}$. But this would be a homomorphism $B_n\to \mathbb Z$ which is surjective upon restricting to $P_n$, which cannot exist.