This has been worked out by Conduché in [Modules croisés généralisés de longueur 2](http://www.sciencedirect.com/science/article/pii/0022404984900343) (JPAA **34** (1984), 155-178), using simplicial group methods.

While it more or less boils down to the stuff described in the answer by Qiaochu, Conduché also describes a nice (2? 3?)-category holding the same information. Its objects are complexes$$G_2\xrightarrow{\partial}G_1\xrightarrow{\partial}G_0$$of $G_0$-groups (composite trivial, $G_0$ acting on itself by conjugation) together with the s. c. Peiffer bracket $\{,\}:G_1\times G_1\to G_2$ satisfying the elaborate but appealing identities
$$
\begin{aligned}
{}^{x_0}\{x_1,y_1\}&=\{{}^{x_0}x_1,{}^{x_0}y_1\}\\
\{\partial x_2,\partial y_2\}&=[x_2,y_2]\\
\partial\{x_1,y_1\}&=x_1y_1x_1^{-1}\left({}^{\partial x_1}y_1\right)^{-1}\\
\{\partial x_2,x_1\}\{x_1,\partial x_2\}&=x_2\left({}^{\partial x_1}x_2\right)^{-1}\\
\{x_1y_1,z_1\}&=\{x_1,y_1z_1y_1^{-1}\}\ \ {}^{\partial x_1}\{y_1,z_1\}\\
\{x_1,y_1z_1\}&=\{x_1,y_1\}\{x_1,z_1\}\{\partial\{x_1,z_1\}^{-1},{}^{\partial x_1}y_1\}.
\end{aligned}
$$
Since both $G_2$ and $G_1$ are nonabelian, it is clear that there will be lots of equivalent nonisomorphic objects, but still this description has its advantages.

Homotopy groups are, as expected, the homology groups of the complex (i. e. $\pi_1$ is the cokernel of $G_1\to G_0$, $\pi_2$ the quotient of the kernel of $G_1\to G_0$ by the image of $G_2\to G_1$, and $\pi_3$ the kernel of $G_2\to G_1$. Thus given $\pi_1$, $\pi_2$, $\pi_3$, to build a model of the above kind one also needs actions of $\pi_1$ on $\pi_2$ and $\pi_3$ as well as the bracket $\pi_2\times\pi_2\to\pi_3$. One then must build a complex as above such that its homology groups are $\pi_1$, $\pi_2$, $\pi_3$, while the actions and the bracket are induced from the structure on the complex.