Let $X$ be the space obtained from the cube $I^3$ by identifying each face with its opposite face via a right-handed one-quarter twist. This is a cell-complex structure on the resulting space:

This is a structure with two 0-cells (the marked vertices represent one equivalence class $p$ and the rest represent an equivalence class $q$), four 1-cells (the labeled edges), three 2-cells (which is clear since we identify the faces in pairs) -- call the front/back $f$, left/right $g$, and bottom/top $h$ -- and one 3-cell $T$. The cellular homology is:

$$0 \rightarrow \la T \ra \xrightarrow{d_3} \la f,g,h \ra \xrightarrow{d_2} \la a,b,c,d \ra \xrightarrow{d_1} \la p, q \ra \rightarrow 0$$

The maps are:

$$d_1 = \left[ \begin{array}{c|cccc} & a & b & c & d \\ \hline p & -1 & 1 & -1 & 1 \\ q & 1 & -1 & 1 & -1 \end{array} \right] \qquad d_2 = \left[ \begin{array}{c|ccc} & f & g & h \\ \hline a & 1 & 1 & 1 \\ b & 1 & 1 & -1 \\ c & 1 & -1 & -1 \\ d & 1 & -1 & 1 \end{array} \right]$$

To determine $d_3$, we compute the degrees of each map in the cellular boundary formula by counting local degrees. First, consider how $T$ is mapped onto the face $f$. The image of the center point of $f$ has a preimage consisting of two points -- the center of the front and back face, respectively. Neighborhoods of these points are mapped homeomorphically onto a neighborhood of the image, so the local degree at each is $\pm 1$. They must have opposite sign, because one neighborhood differs from the other by a reflection and a rotation, and this would multiply the degree by $-1$ and $1$ respectively, yielding an opposite sign. Then, the degree of the map is $1 - 1 = 0$. We get the same thing for the other faces, so each coefficient in the sum in the cellular boundary formula is zero, and $d_3 = 0$.

Let's compute homology. We have:

$$d_1 \xrightarrow{SNF} \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \qquad d_2 \xrightarrow{SNF} \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right]$$

Now that we have the maps explicitly and their Smith forms, the homology is just a matter of tabulation:

$$H_0(X) \cong \mathbb{Z} \qquad H_1(X) = \mathbb{Z}_2 \oplus \mathbb{Z}_2 \qquad H_2(X) = 0 \qquad H_3(X) = \mathbb{Z}$$