Oops ! Mea culpa ! Indeed, the decomposition of $\mathbb{S}^3/\mathcal{I}$ is not regular indeed, but the $\mathcal{I}$-equivariant one we found on the sphere is. More zeros appear when applying the functor $-\otimes_{\mathbb{Z}[\mathcal{I}]}\mathbb{Z}$ to the equivariant chain complex to obtain the non-equivariant one on the orbit space. Thank you very much @HenrikRüping for your correction !

You're almost right ! For a regular CW-decomposition and with integral coefficients, the homology chain complex must only involve differentials with entries $\pm1$ or 0. As I have mentioned, the differentials in our complex do respect this condition. In particular, the third differential is zero and has source $\mathbb{Z}$, so $H_3=\mathbb{Z}$ as expected. The middle differential $d_2 : \mathbb{Z}^5 \to \mathbb{Z}^5$ is invertible, so $H_1=H_2=0$ and the first differential $d_1 : \mathbb{Z}^5\to\mathbb{Z}$ is zero, so $H_0=\mathrm{coker}(d_1)=\mathbb{Z}$.