The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A_q$, a <i>simple</i> one. Let $K=\mathcal A(1).$ Then

$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad \text{($m$ times)} \qquad (*)$

Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F_n) \to A_q$ which doesn't vanish at $\alpha$.

Then <p style="text-indent: 20px;"> $  A_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K).  $ </p> It follows that <p style="text-indent: 20px;"> $  \Delta( [K,K,\ldots,K])=A_q  $ </p> and by $(*)$ <p style="text-indent: 20px;"> $  \Delta( \mathcal A(m))=A_q.  $ </p> for every $m \ge 1.$