My question is just as in the box. Is every smooth projective toric variety diffeomorphic to a quotient of $\prod_i S^{n_i} \times T^k$ (I know torus is a onesphere but I just wanted to make clear I allow this) by a free torus action? If not which ones can be realized like this? Maybe it can be proven using the Geometric Invariant theory construction. Namely, when is the fiber of the corresponding torus action on $\mathbb{C}^l$ diffeomorphic to the one of the above spaces?
Let us consider the case of toric varieties of real dimension $4$ and prove they can not be represented as such a quotient unless they have second Betti number $1$ or $2$. Proof. Let us introduce some notations. Let $B$ be the toric manifold of real dimension $4$, $n=b_2(B)$. Denote by $E$ the product $\Pi_i S^{n_i}\times T^k$. Let $k_2=b_2(E)$, $k_3=b_3(E)$. Finally, denote by $m$ the dimension of the torus that is acting on $E$. Suppose that $T^m$ is acting freely on $E$ and $B=E/T^m$. Then we have the following obvious relation on dimensions: $$4=dim(E)m\ge k+2k_2+3k_3m$$ We will explain now that we must have $n\le 2$ (recall $n=b_2(B)$). For this purpose we will consider the long exact sequence of homotopy groups for the fibration $E\to B$. $$0\to \pi_3(E)\to \pi_3(B)\to 0 \to \pi_2(E)\to \pi_2(B)\to \mathbb Z^m\to \mathbb Z^k\to 0$$ Since $\pi_2(B)=\mathbb Z^n$, from the second half of the sequence we get $$k_2+nm+k=0$$ substituting this in the inequality on dimensions we get $$4\ge 3k_2+3k_3n$$ To prove finally that $n\le 2$ we use the classical statement that $\pi_3(B)$ contains subgroup $\mathbb Z^{(n^2+n)/21}$. It follows that $$rk (\pi_3(E))=k_3+k_2\ge (n^2+n)/21$$ so $$4\ge 3((n^2+n)/21)n$$ 

