Let $A'$ and $B'$ be the classical adjoints of $A$ and $B$. Write
$$S(A,B)=(B+B')A^2B-AB^2(A+A')$$ $$T(A,B)=BA^2(B+B')-(A+A')B^2A$$ $$U(A,B)=B'A^2B-AB^2A'$$ $$V(A,B)=BA^2B'-A'B^2A$$
Then:
For any $n$ we have (trivially) $$S(A,B)-T(A,B)=U(A,B)-V(A,B)$$
For $n=2$ (but not otherwise), $S(A,B)$ is equal to your left-hand side and $T(A,B)$ is equal to your right-hand side.
For $n=2$ (but not otherwise) it's easy to check that $U(A,B)=V(A,B)$.
The proof of (2) above relies on the fact that, for $n=2$ (but not otherwise) we have $B+B'=trace(B)$, and therefore $B+B'$ commutes with everything. There's no good generalization of this to $n>2$.
With a little more patience and/or cleverness, one might hope to juggle the expressions for $S,T,U$ and $V$ a little so that 1) and 2) remain true as stated, while 3) becomes true for every $n$. This would essentially imply that your identity is equivalent to $B+B'=tr(B)$ and explain why we shouldn't expect it to generalize past $n=2$.