Exact sequence of $n$th powers of abelian groups
By Ycor's arguments, it seems we can reduce to $p$-groups.
Assume $A$ and $B$ are finite $p$-groups. The $p$-rank of $A$ must equal the $p$-rank of $B$ or else the right exact sequence $(C/pC) \to (A/pA)^n\to (B/pB)^n \to 0$ would imply that the $p$-rank of $C$ is at least $n$ and hence that $C$ has at least $n$ generators.
But given that the $p$-rank of $A$ equals the $p$-rank of $B$, the same right exact sequence implies that $(A/pA)^n\to (B/pB)^n$ is an isomorphism, and so the sequence $0 \to C \to (pA)^n \to (pB)^n \to 0$ exists and is exact. So we might as well work with $pA$ and $pB$.
We can now apply induction on the cardinality of $A$ and $B$ to reduce to the base case when $A=B=0$, which implies $C=0$.
Rather than using Ycor's reduction, one can instead rephrase the inductive argument as a proof that $(A / mA)^n \to (B / m B)^n$ is an isomorphism for all natural numbers $m$, which implies that $A^n \to B^n$ is an isomorphism by a standard lemma.
Let me try to write carefully the argument for Noetherian rings. Let $C \to A^n \to B^n$ be an exact sequence of $R$-modules for a Noetherian ring $R$ where $C$ has fewer than $n$ generators. Our goal will be to show that $C$ vanishes. Let $\mathfrak m$ be a maximal ideal of $R$.
Consider the right exact sequence $(C/\mathfrak m C) \to (A/\mathfrak m A)^n\to (B/\mathfrak m B)^n \to 0$. We have $n \dim (A/\mathfrak m A) \leq \dim (C/\mathfrak m C) + n \dim(B/\mathfrak m B)$ and because every dimension is an integer and $\dim (C/\mathfrak m C) $, this implies that $\dim (A/\mathfrak m A) = \dim (B/\mathfrak m B)$ and hence the right map is an isomorphism, so $0 \to C \to (\mathfrak m A)^n \to (\mathfrak mB)^n \to 0$ is an exact sequence.
Iterating this argument, we see that for all $k$, $(A/\mathfrak m^k A)^n \to (B/\mathfrak m^k B)^n$ is an isomorphism. Hence the induced map from the $\mathfrak m$-adic completion of $A$ to the $\mathfrak m$-adic completion of $B$ is an isomorphism. Because $\mathfrak m$-adic completion of a Noetherian local ring is an exact functor on finitely generated modules, this implies the $\mathfrak m$-adic completion of $C$ is zero. In particular, $C/\mathfrak m C$ is zero. Because this holds for all maximal ideals $\mathfrak m$, and $C$ is a finitely-generated module, $C$ vanishes.
Here's a partial reduction; I write it here since I was too sloppy in the comments.
Lemma Let $R$ be a noetherian (commutative) ring. Let $$0\to M\to N\to Q\to 0$$ be a short exact sequence of f.g. modules with $M\neq 0$. Then there exists an artinian local ring quotient $R'$ of $R$ such that the induced surjective map $N\otimes_R R'\to Q\otimes_R R'$ is non-injective.
Indeed, let $m\in N$ be a nonzero element in the kernel of $N\to Q$. Since any f.g. module over a noetherian ring is residually of finite length ($\star$), there exists an ideal $I$ of $R$ such that $R'=R/I$ is of finite length and $m\notin IN$. This proves the claim at least with $R'$ artinian (maybe not local), but local follows since $R'$ is a direct product of finitely many local rings.
This reduces the question from noetherian rings to its local artinian quotients (this, in the case of $R=\mathbf{Z}$, reduces to finite $p$-groups).
($\star$) Let $W$ be a maximal submodule of $N$ not containing $m\neq 0$. It's enough to prove that $N/W$ has finite length. Indeed, in the quotient $N/W$, the intersection of nonzero submodules is nonzero. But if $N/W$ has infinite length, it has some non-maximal associated prime $P$, but then in $R/P$ the intersection of nonzero ideals is zero (e.g., using powers of a single ideal), contradicting that $R/P$ is isomorphic to a submodule of $N/W$.
For people not familiar to commutative algebra, here's the particular case of abelian groups.
Lemma' Let $$0\to M\to N\to Q\to 0$$ be a short exact sequence of f.g. abelian groups with $M\neq 0$. Then there exists a prime power $n\ge 1$ such that the induced surjective map $N/nN\to Q/nQ$ is non-injective.
Indeed, let $m\in N$ be a nonzero element in the kernel of $N\to Q$. There exists $n\ge 1$ such that $m\notin nN$. This proves the lemma at least with $n\ge 1$ (maybe not prime power), and then passing to some $p$-power is immediate by the Chinese remainder theorem.