Due to better editing facilities I post this as an answer, though it's rather a comment. My motivation for beliving the long sequence might be exact was the (unconditional) exactness of the sequence in remark 1. But when applied to the short exact sequence $$ 0 \to \Omega(B) \to P \to B \to 0$$ with $P$ projective, the two long sequences show a significant difference. For, since $\Omega^n(P)$ is again projective we have $[-,\Omega^n(P)] = 0$ and the sequence in remark 1 just yiels the tautological $$[A,\Omega^n(B)] \cong [A, \Omega^{n-1}(\Omega(B))],$$ while the sequence in question yiels the sequence $$0 \to [\Omega^n(A),B] \to [\Omega(\Omega^n(A)), \Omega(B)] \to 0.$$ I see no reason, why this sequence should be exact in general. Therefore, I guess, one needs further assumptions, like $\mathcal{A}$ being Frobenius as observed by Sasha, in order to make the questionable long sequence exact.

Due to better editing facilities I post this as an answer, though it's rather a comment. My motivation for beliving the long sequence might be exact was the (unconditional) exactness of the sequence in remark 1. But when applied to the short exact sequence $$ 0 \to \Omega(B) \to P \to B \to 0$$ with $P$ projective, the two long sequences show a significant difference. For, since $\Omega^n(P)$ is again projective we have $[-,\Omega^n(P)] = 0$ and the sequence in remark 1 just yields the tautological $$[A,\Omega^n(B)] \cong [A, \Omega^{n-1}(\Omega(B))],$$ while the sequence in question yields the sequence $$0 \to [\Omega^n(A),B] \to [\Omega(\Omega^n(A)), \Omega(B)] \to 0.$$ I see no reason, why this sequence should be exact in general. Therefore, I guess, one needs further assumptions, like $\mathcal{A}$ being Frobenius as observed by Sasha, in order to make the questionable long sequence exact.