Quotient of $ S^3 \times S^3$ by a free torus action.

This began as a comment on Nick L's answer, but got too long.

Actually, I think TuoTuo is correct that the quotient is $S^2\times S^2$ independent of $n$ and $k$.

To see this, let $x = a + bj\in S^3$ with $a,b\in \mathbb{C}$ and both $a$ and $b$ non-zero. Under the $w$ action, the equivalence class of $x$ is all quaternions of the form $x\overline{w}$.

The $z$ action on $[x]\in S^2 = S^3/S^1$ is given by $z\ast[x] = [z^{n+2k} x]$.

Proposition: For $z\in S^1$, $z\ast[x] = [x]$ iff $z$ is a $2(n+2k)$th root of $1$.

Proof: First, if $z$ is such a root of $1$, then $z^{n+2k} = \pm 1$, so commutes with $x$. If we set $w = z^{n+2k}$, then $z^{n+2k}\overline{w} = z^{n+2k}w = 1$ and so $$z\ast[x] = z\ast[x\overline{w}] = [z^{n+2k} x \overline{w}] =[x].$$

Conversely, if there is a $w\in S^1$ with the property that $z^{n+2k}\, x\, \overline{w}= x$, then we must have $z^{n+2k}\, \overline{w}\, a = a$ and $z^{n+2k}\, w\, b = b$ (where the scond equality uses the fact that $j\overline{w} = wj$). Since we are assuming $a$ and $b$ are both non-zero, this forces $z^{n+2k}\, \overline{w} = 1$ and $z^{n+2k}\, w = 1$. Multiplying these two equations together gives $z^{2(n+2k)} = 1$ as claimed. $\square$

It follows from this proposition that the $S^1$ action on $S^2$ rotates $S^2$ around $2(n+2k)$ times. Since this an even number, it is homotopically trivial. This, then, implies the quotient is the trivial bundle, so is diffeomorphic to $S^2\times S^2$.

Actually, this answer was wrong, but it seems Jason's answer continues from the observations here, so I don't delete. We proceed by quotienting the two factors of $S^{1} \times S^{1}$ "one-by-one". Firstly note that the action of $\{1\} \times S^{1}$ on $S^{3} \times S^{3}$ by $\{1\} \times S^{1}$ is just the standard Hopf fibration on the second factor and acts trivially on the first factor, hence the quotient $S^{3} \times S^{3} / \{1\} \times S^{1}$ is diffeomorphic to $S^{3} \times S^{2}$ (actually the "direction" of this $S^{1}$-action is the opposite of the standard Hopf fibration since $\bar{w}$ appears, but the orbits are the same, hence the quotient is $S^{2}$- I can make it more explicit if you want).

Note now, that since $S^{1} \times S^{1}$ is abelian the action of $S^{1} \times \{1\}$ descends to a free action on the quotient $S^{3} \times S^{2}$. Lastly, one may see that the final quotient is an $S^{2}$-bundle over $S^{2}$ (with the projection being induced by the projection to the left factor). See Jason DeVito's answer-infact the fibration is trivial.