This is very computable. Let $G^{(k)}$ be the lower central series of $G$, ie $G^{(1)}=G$ and $G^{(k+1)} = [G^{(k)},G]$. There are algorithmic ways to compute the quotients $G^{(k)}/G^{(k+1)}$ (eg using the Fox free differential calculus -- see Fox's series of papers on the free differential calculus for the details). The direct sum
$$\oplus_{k=1}^{\infty} G^{(k)} / G^{(k+1)}$$
has the structure of a graded Lie algebra with the Lie bracket induced by conjugation (this is explained in many places -- I recommend the last chapter of Magnus-Karass-Solitar's book on combinatorial group theory or Serre's book "Lie Algebras and Lie Groups"). . This Lie algebra is generated by the degree 1 piece, namely $G^{(1)} / G^{(2)} \cong G^{ab}$. The degree 2 piece is a quotient of $\wedge^2 G^{ab}$ by some subgroup $R$. I claim that understanding $R$ is exactly what you need to know to understand the kernel of the cup product map. Namely, we have a surjection
$$\wedge^2 G^{ab} \rightarrow \wedge^2 G^{ab} / R$$
and thus a dual injection
$$(\wedge^2 G^{ab} / R)^{\ast} \hookrightarrow \wedge^2 (G^{ab})^{\ast}.$$
The image of this injection is exactly the kernel of the cup product map.

Let me sketch a proof. To simplify things, let's assume that everything in site is torsion-free (it will simplify our statements). Set $H = H_1(G)$ and $H^{\ast} = H^1(G) = Hom(H,\mathbb{Z})$. The above will allow you to compute the kernel of the cup product map $\wedge^2 H^{\ast} \rightarrow H^2(G)$ as follows. Consider the short exact sequence

$$1 \longrightarrow G^{(2)} \longrightarrow G \longrightarrow H \longrightarrow 1.$$

There is an associated 5-term exact sequence in group cohomology which takes the form

$$0 \longrightarrow H^1(H) \longrightarrow H^1(G) \longrightarrow (H^1(G^{(2)}))^H \longrightarrow H^2(H) \longrightarrow H^2(G).$$

Now, the map $H^1(H) \rightarrow H^1(G)$ is an isomorphism. Also, $H^2(H) = \wedge^2 H^{\ast}$, and the map $H^2(H) \rightarrow H^2(G)$ is easily seen to be the cup product map. What we deduce is that we have an exact sequence

$$0 \longrightarrow (H^1(G^{(2)}))^H \longrightarrow \wedge^2 H^{\ast} \longrightarrow H^2(G).$$

In other words, the kernel of the cup product map is the subgroup $(H^1(G^{(2)}))^H$ of $\wedge^2 H^{\ast}$.

Let us now interpret this subspace. It is easiest to dualize. The dual of the above inclusion is the surjection

$$H_2(H) \rightarrow (H_1(G^{(2)}))_H.$$

Now, $H_1(G^{(2)})$ is just $G^{(2)} / [G^{(2)},G^{(2)}]$, and we are killing off the action of $H$, which is the same as killing off the conjugation action of $G$. In other words, we have an isomorphism

$$(H_1(G^{(2)}))_H \cong G^{(2)} / [G,G^{(2)}] = G^{(2)} / G^{(3)}.$$

The desired claim is an immediate consequence.