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Let $G$ be a group. My question is whether the following equivalence holds:

$$ \pi_{2}(\mathbf{B}^{2}G)\simeq ab(G) $$

Obviously higher homotopy groups are all abelian, but what happens when you suspend a non-abelian classifying space? My intuition says that these spaces must have the abelianization of the group as their only homotopy group, concentrated in degree two. I don't know how I would go about showing this, and for some reason I can't find any resources online that state this. If my intuition is incorrect, how instead should I be thinking about these spaces?

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    The double delooping of a group doesn't make sense unless it's abelian. Delooping is not suspension. – 2017-01-19
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    Okay. How does the suspension of a non-abelian group's classifying space relate to its abelianization, if at all? – 2017-01-19

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$BG$ is a connected space with the property that $H_1(BG) \cong G/[G, G]$ is the abelianization of $G$ by Hurewicz. Suspending $BG$ produces a simply connected space whose homology groups are those of $BG$ shifted up by one, hence $H_2(SBG) \cong G/[G, G]$ and $H_2(SBG) \cong \pi_2(SBG)$ by Hurewicz again.

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I don't follow part of your question, as in your displayed formula, but if $G$ is a group then we can form $K(G,1)= BG$, say, and something complete can be said about the homotopy $3$-type of the suspension $Y= SK(G,1)$, but in general the question is complicated. For example, if $G$ is the group of integers, then $S^1$ is a model of $K(G,1)$ and the suspension of $S^1$ is of the homotopy type of $S^2$ whose homotopy groups have not been completely described.

However a description of the $3$-type of $Y$ is given in my paper with J.-L. Loday in Topology 26 (1987) 311-335, of which here is a version. You need some new concepts, e.g. of "crossed square" and the one in particular in this case is partly described as

$$\begin{matrix} G \otimes G &\to & G \\ \downarrow && \downarrow \\ G & \to & G\end{matrix}$$ where $G \otimes G$ is the nonabelian tensor square of the group $G$. The key property of this construction is that there is a morphism of groups $\kappa: G \otimes G \to G $ and a function $h: G \times G \to G\otimes G $ such that $\kappa h$ is the commutator map. Then $\pi_3SK(G,1)$ is isomorphic to the kernel of the morphism $ \kappa $. The cokernel of $\kappa $ is $G$ made abelian, and this is isomorphic to $\pi_2(SK(G,1))$.

There is a lot more information on the nonabelian tensor product on this bibliography, including some expositions, and plenty of group theoretic results.