This is a partial answer.
For any abelian group $G$ and integer $n \geq 1$, there is a CW complex $M(G, n)$ such that
$$\tilde{H}_i(M(G, n)) = \begin{cases}
G & i = n\\
0 & i \neq n
\end{cases}$$
called a Moore space. Moreover, for $n > 1$, we can take these spaces to be simply connected. See Example $2.40$ of Hatcher's Algebraic Topology.
Using the fact that $H_n(\bigvee_{\alpha}X_{\alpha}) = \bigoplus_{\alpha}H_n(X_{\alpha})$, we see that for a given sequence of abelian groups $\{G_n\}_{n=1}^{\infty}$, the CW complex $X = \bigvee_{n=1}^{\infty}M(G_n, n)$ has the property that $H_i(X) = G_i$.
However, all that can be said of $\pi_1(X)$ is that its abelianisation is $G_1$. There are many different groups with the same abelianisation, so $\pi_1(X)$ may not be the group you wanted it to be.
If one could construct the Moore spaces $M(G, 1)$ with a given fundamental group (which must satisfy the necessary condition that its abelianisation is $G$), then the above space would have the desired properties. However, I don't know if this has been done.
Added Later: I asked a separate question about this final point (whether one could construct a Moore space $M(G, 1)$ with given fundamental group) here. It turns out it is not always possible.
I don't think that there is a better description of the space than the one i gave in the comments. Let me give slightly more details.
There is a CW-structure for $S^1\times S^1\times S^1$ with a single $0$-cell, three $1$-cells, three $2$-cells, and a single $3$-cell. The picture of this is the one from the question. For more details, see this post. We then get that the $1$-skeleton is $S^1\vee S^1\vee S^1$. The $2$-cells are attached via the words $aba^{-1}b^{-1}$, $aca^{-1}c^{-1}$, and $bcb^{-1}c^{-1}$ where $a$, $b$, and $c$ denotes the three $1$-cells. The $2$-skeleton can also be seen as hollowing out the cube from the question. Note that removing a point from $S^1\times S^1\times S^1$ results in a space that deformation retracts to the $2$-skeleton. This is analogous to the case of the $2$-torus which you have drawn. Let us denote this space with $X$.
Now let me give another geometric description of $X$. Let us attach the $2$-cells one at a time. Attaching the $2$-cell corresponding to the word $aba^{-1}b^{-1}$ yields the space $S^1\vee (S^1\times S^1)$ as seen in the picture:
Then one glues another $2$-cell via the word $aca^{-1}c^{-1}$. This yields a wedge of two tori with two $1$-cells from their $1$-skeleta identified. Attaching the last $2$-cell yields a wedge of three tori such that each torus has the two $1$-cells identified with a $1$-cell from each of the other tori respectively. This yields another description of $X$. Let $(S^1\times S^1)_i$ for $i=1,2,3$ be three tori. Then the space can be described as the quotient
$$
X=\frac{(S^1\times S^1)_1\vee (S^1\times S^1)_2 \vee (S^1\times S^1)_3}{(z,*)_1\sim (z,*)_2,\, (*,z)_1\sim(*,z)_3,\, (*,z)_2\sim(z,*)_3}
$$
where $*$ denotes the basespoint of the circle. (For example pick $*=(1,0)\in S^1$).
Best Answer
Let $X$ be your space, and note that it can be realized as a quotient of $I^2$. In particular, the usual structure of the torus is a quotient map $q:I^2 \to I^2/\sim$ by taking $(x,0) \sim (x,1)$ and $(0,y) \sim (1,y)$.
There is another quotient now $p:I^2 \to I^2/\sim$ by taking $(x_1,0) \sim (x_2,0)$ and $(x_1,1) \sim (x_2,1)$. Your space is done by doing $q$ first and then $p$ (or technically the map induced by $p$.)
However, the situation is clearer if you do $p$ first and then $q$.
Pinching two sides of the square down to a point gives a structure of $2$ zero cells, and two $1$-cells joining them, and filling this with a $2$-cell. The induced bap by $q$ glues these two $1$ cells together giving a sphere, but we still have to identify the $0$-cells, so this is a sphere with two points identified.
There is an excellent illustration that a sphere with two points identified is homotopically just $S^2 \vee S^1$ here.
Now one can apply siefert-Van kampen to obtain the result.