how many non equivalent $\mathbb{T}^n$ bundles are there over $\mathbb{S}^n$? Intuitively, maybe you have $\mathbb{Z}$  in inequivalent bundles for each $\mathbb{S}^1$ component as you can imagine taking the torus product an interval and gluing together the ends of the interval with a "twist". But this might not be true and I don't know how to go about proving it if it is....
You have the exact sequence $0\rightarrow \mathbb{Z}^n\rightarrow \mathbb{C}^n\rightarrow\mathbb{T}^n\rightarrow 1$ which induces an isomorphism (Cech cohomology) $H^2(M,\mathbb{Z}^n)\rightarrow H^1(M,\mathbb{T}^n)$. Take $M=S^n$, $H^2(M,\mathbb{Z}^n)=0$ if $n\neq 2$ so for $n\neq 2$ the bundle are trivial.
You have the exact sequence $0\rightarrow \mathbb{Z}^n\rightarrow \mathbb{C}^n\rightarrow\mathbb{T}^n\rightarrow 1$ which induces an isomorphism (Cech cohomology) $H^2(M,\mathbb{Z}^n)\rightarrow H^1(M,\mathbb{T}^n)$. Take $M=S^n$, $H^2(M,\mathbb{Z}^n)=0$ if $n\neq 2$ so for $n\neq 2$ the bundle are trivial.