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 simple one (see his comment above). Let $K=mathcal A(1).$ Then
$mathcal A(m) ge [K,K,ldots,K]=[..[K,K],..,K]qquad (mquad times) qquad (*)$
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
$$
A_q =mathrm{NormalClosure}(Delta(alpha))=Delta([K,K])=Delta(K).
$$
It follows that
$$
Delta( [K,K,ldots,K])=A_q
$$
and by $(*)$ $Delta( mathcal A(m))=A_q $ for every $m ge 1.$
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