Here are some thoughts; I don't know if they'll add up to a satisfying answer. Let $X$ be $(n-1)$-connected. We have a Hurewicz map $\pi_n(X) \to H_n(X)$ given by applying $H_n$ to maps $S^n \to X$ and we want to know that it's an isomorphism (or abelianization if $n = 1$). Applying $\text{Hom}(-, A)$ for an arbitrary abelian group $A$, the Yoneda lemma shows that this is equivalent to knowing that the corresponding map
$$\text{Hom}(H_n(X), A) \to \text{Hom}(\pi_n(X), A)$$
is always an isomorphism. Now, by the universal coefficient theorem $\text{Hom}(H_n(X), A)$ can be identified with $H^n(X, A)$, so this map can be thought of as the map
$$H^n(X, A) \to \text{Hom}(\pi_n(X), A)$$
given by applying $\pi_n$ to maps $X \to B^n A$, once we've shown that the classifying space $B^n A \cong K(A, n)$ exists and represents cohomology (I don't know off the top of my head whether there's a clean way to do this that avoids the Hurewicz theorem).
The nice thing about having massaged the statement to this form is that now it is entirely a statement about computing homotopy classes of maps and we can hope for a reasonably conceptual $\infty$-categorical way to understand, if not non-circularly prove, it. My understanding would use the following: there is a functor $\tau_{\le n}$ sending a space to its $n$-truncation, which is left adjoint to the inclusion from $n$-truncated spaces (spaces with vanishing $\pi_k, k \ge n+1$) into spaces, and which correspondingly has the universal property that
$$\text{Map}(\tau_{\le n} X, Y) \cong \text{Map}(X, Y)$$
for any $n$-truncated space $Y$. Now, $B^n A$ is $n$-truncated, so setting $Y = B^n A$ and $X$ as above gives
$$\text{Map}(\tau_{\le n} X, B^n A) \cong \text{Map}(X, B^n A)$$
or, restated in terms of cohomology,
$$H^n(\tau_{\le n} X, A) \cong H^n(X, A).$$
In other words, $H^n$ only depends on the $n$-truncation of $X$. This is useful because the assumption that $X$ is $(n-1)$-connected implies that $\tau_{\le n} X$ has only a single nonzero homotopy group, and hence it can be identified with $B^n \pi_n(X)$. Now we've reduced to showing that
$$\text{Map}(B^n \pi_n(X), B^n A) \cong \text{Hom}(\pi_n(X), A)$$
which at least morally follows from taking loop spaces $n$ times and using that abelian groups fully faithfully embed into $n$-fold loop spaces, $n \ge 1$ (for $n = 1$ and $\pi_1$ nonabelian, the same statement for groups). I also don't know whether there's a clean proof of this that avoids the Hurewicz theorem!