This is an extended-comment, but I think no:
In homology, we know that $H_1$ is abelianization, $H_2$ is Schur multiplier, but $H_{n\ge 3}$ is ???
In cohomology, $H^1(G;A)$ is split extensions, and this fits well with $Hom(G,A)$ in the UCT. As you mention, $H^2(G,A)$ corresponds to group extensions of $G$ by $A$, and this fits well with $Hom(H_2G,A)$ in the UCT. But as we see, this all ties in to knowing $H^n(G;A)$ and $H_n(G;A)$ very well...
If we look at $H^3(G;A)$ we get crossed module extensions $0\to A\to N\to E\to G\to 0$, and these are cumbersome. We have no nice interpretation for $H^n(G;A)$ for $n>3$, except more crazy-looking exact sequences. This is why I don't expect a "nice" map (i.e. "interpretation" of it) in the UCT to arise.
I'll start by describing the notation that I'll use.
I'll think of elements of $\mathbb{Z}^\mathbb{N}$ as infinite column vectors
$${\bf x}=\begin{pmatrix}
x_0\\x_1\\x_2\\\vdots
\end{pmatrix}$$
of integers, with rows indexed by $\mathbb{N}$.
The $i$th "unit vector" will be denoted by ${\bf e(i)}$. I.e., $e(i)_i=1$ and $e(i)_j=0$ for $j\neq i$.
As described in the question, Specker's result implies that endomorphisms of $\mathbb{Z}^\mathbb{N}$ are of the form ${\bf x}\mapsto A{\bf x}$ where
$$A=\begin{pmatrix}
a_{00}&a_{01}&a_{02}&\cdots\\
a_{10}&a_{11}&a_{12}&\cdots\\
a_{20}&a_{21}&a_{22}&\cdots\\
\vdots&\vdots&\vdots&\ddots
\end{pmatrix}$$
is an infinite matrix of integers, with rows and columns indexed by $\mathbb{N}$, that is row-finite (i.e., each row has only finitely many non-zero entries).
Now suppose that $\alpha:\mathbb{Z}^\mathbb{N}\to\mathbb{Z}^\mathbb{N}$ is a surjective group homomorphism, given by a row-finite matrix $A$.
To prove that $\alpha$ is split, we need to find a right inverse. In fact, it suffices to find another row-finite matrix $B$ such that $AB$ is lower unitriangular (i.e., diagonal entries are all $1$ and entries above the diagonal are all $0$), since such a matrix is invertible.
Fix $n\geq0$, and decompose $\mathbb{Z}^\mathbb{N}=G_n\oplus H_n$, where $G_n\cong\mathbb{Z}^n$ is the subgroup consisting of those ${\bf a}$ with $a_i=0$ for $i\geq n$, and $H_n\cong\mathbb{Z}^\mathbb{N}$ is the subgroup consisting of those ${\bf a}$ with $a_i=0$ for $i<n$.
Since $\alpha$ is surjective, $\mathbb{Z}^\mathbb{N}=\alpha\left(G_n\right)+\alpha\left(H_n\right)$.
Consider the quotient maps
$$\mathbb{Z}^\mathbb{N}\stackrel{\theta}{\to} \mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\stackrel{\varphi}{\to} \left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)/T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right),$$
where $T(X)$ denotes the torsion subgroup of an abelian group $X$.
Since $\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)/T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$ is a finite rank free abelian group, using Specker's result again implies that $\varphi\theta({\bf e(i)})=0$, or equivalently $\theta({\bf e(i)})\in T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$, for all but finitely many $i$.
Since $T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$ is finite, for all but finitely many $i$ with $\theta({\bf e(i)})\in T\left(\mathbb{Z}^\mathbb{N}/\alpha\left(H_n\right)\right)$, there is $i'>i$ with $\theta({\bf e(i')})=\theta({\bf e(i)})$, or equivalently ${\bf e(i)}-{\bf e(i')}\in\alpha(H_n)$.
So we can choose a sequence $0=t_0<t_1<t_2<\dots$ of integers such that for every $i\geq t_n$ there is some $i'>i$ and some ${\bf b(i)}\in H(n)$ with ${\bf e(i)}-{\bf e(i')}=\alpha\left({\bf b(i)}\right)$. For each $i$, do this with the largest $n$ such that $i\geq t_n$, and let $B$ be the matrix whose $i$th column is ${\bf b(i)}$.
Then $B$ is row-finite, and $AB$ is lower unitriangular.
Best Answer
Nice question. For a prime $p$ let $\mathbb{Z}_{(p)} = \lbrace \frac{a}{b}\in \mathbb{Q}\mid p \nmid b\rbrace$ and $\mathbb{Z}[p^{-1}] = \lbrace \frac{a}{p^n}\in \mathbb{Q}\mid n \ge 0 \rbrace$. Then
$$A := \lbrace (x,y) \in \mathbb{Q} \times \mathbb{Z}[p^{-1}] \mid x-y \in \mathbb{Z}_{(p)} \rbrace$$ has the desired non-split extension. Informal $A$ consists of all pairs $(x,y) \in \mathbb{Q} \times \mathbb{Q}$ where $y$ is the $p$-part in the partial fraction decomposition of $x$ (up to an integer summand).
Proof: Let $\rho: A \to \mathbb{Q}$ be projection onto the first factor and $i: \mathbb{Z} \hookrightarrow A$ inclusion into the second factor. By partial fraction decomposition of the rationals, $\rho$ is surjective and $$\ker(\rho)=0 \times \big(\mathbb{Z}[p^{-1}] \cap \mathbb{Z}_{(p)}\big) = 0 \times \mathbb{Z} = \operatorname{im}(i).$$
Next, let $j: \mathbb{Q} \to A$ be a splitting hom. of $\rho$. Composing $j$ with the projection onto the second factor yields a hom. $f: \mathbb{Q} \to \mathbb{Z}[p^{-1}] \le \mathbb{Q}$. Since each endomorphism of $\mathbb{Q}$ is multiplication with some $q \in \mathbb{Q}$ we have $f(x)=qx \in \mathbb{Z}[p^{-1}]$ for all $x \in \mathbb{Q}$ which is only possible for $q=0$. Hence $j(x)=(x,0) \in A$ for all $x \in \mathbb{Q}$. Setting $x=1/p$ we obtain $1/p \in \mathbb{Z}_{(p)}$ which is the desired contradiction. QED