Given a smooth manifold $M$, we define the differentiable structure on $TM$ in the usual way.
I would like to know examples of two smooth manifolds which are non-diffeomorphic, but their tangent bundles are.
Which is the smallest dimension in which one can find such examples?
What if I ask the same question for $k$ pairwise non-diffeomorphic manifolds?
Can we have $k=\infty$?
Here are examples of non-diffeomorphic closed manifolds with diffeomorphic tangent bundles:
3-dimensional lens spaces have trivial tangent bundles, which are diffeomorphic if and only if the lens spaces are homotopy equivalent, e.g. $L(7,1)$, $L(7,2)$ are not homeomorphic, but their tangent bundles are diffeomorphic. This follows from proofs in [Milnor, John, Two complexes which are homeomorphic but combinatorially distinct. Ann. of Math. (2) 74 1961 575--590].
Tangent bundle to any exotic $n$-sphere is diffeomorphic to $TS^n$ as proved in [De Sapio, Rodolfo, Disc and sphere bundles over homotopy spheres, Math. Z. 107 1968 232--236].
In dimensions $n\ge 5$ one can attack this question via surgery theory. For example, let $f:N\to M$ be a homotopy equivalence of closed $n$-manifolds that has trivial normal invariant (which is a bit more than requiring that $f$ preserves stable tangent bundle). Multiply $f$ by the identity map of $(D^n, S^{n-1})$, where $D^n$ is the closed $n$-disk. Then Wall's $\pi-\pi$ theorem implies that $M\times D^n$ and $N\times D^n$ are diffeomorphic, so if tangent bundles of $M, N$ are trivial, this gives manifolds with diffeomorphic tangent bundles.
To illustrate the method in 3, here is a particular example of infinitely many non-homeomorphic closed manifolds with diffeomorphic tangent bundles. Fix any closed $(4r-1)$-manifold $M$ where $r\ge 2$ such that $TM$ is stably trivial and $\pi_1(M)$ has (nontrivial) elements of finite order. Then results of Chang-Weinberger imply that there are infinitely many closed $n$-manifolds $M_i$ that are simply-homotopy equivalent to $M$ and such that the tangent bundles $TM_i$ are all diffeomorphic (I am not quite sure how to get them be diffeomorphic to $M\times\mathbb R^n$ even though this should be possible). I know how to deduce this from [On invariants of Hirzebruch and Cheeger-Gromov, Geom. Topol. 7 (2003), 311--319].
I have been thinking extensively of related issues, so you might want to look at my papers at arxiv, e.g. this one.
Another good $2$-dimensional example is torus punctured once and sphere punctured three times. These become diffeomorphic when crossed with $\mathbb R$, and they have trivial tangent bundles. More generally, if you puncture a closed genus $g$ surface $n>0$ times then the product with $\mathbb R$ depends only on $2g+n$, so in dimension $2$ you can get arbitrarily large finite $k$.
But maybe you wanted a compact (without boundary) example?
EDIT In the spirit of Igor's (3), suppose we have closed smooth $n$-manifolds $M_1$ and $M_2$ with rank $k$ vector bundles $\xi_1$ and $\xi_2$, and we want to know whether the two $n+k$-manifolds $D(\xi_i)$ (closed disk bundles) are diffeomorphic. There is the necessary condition: there is a simple homotopy equivalence $f:M_1\to M_2$ such that $TM_1\oplus\xi_1$ is isomorphic (as vector bundle) to $f^*(TM_2\oplus\xi_2)$. In some cases this is sufficient. If $k>n$ then $f$ gives an embedding of $M_1$ in $D(\xi_2)$, and the normal bundle is isomorphic to $\xi_1$, so $M_1$ embeds in $D(\xi_2)$ with tubular neighborhood $D(\xi_1)$. The h-cobordism theorem shows that the outside of this is a collar. This works as long as $n+k$ is not too small. You don't need the homotopy equivalence to be simple if you just want a diffeomorphism of open disk bundles. In the case $k=n>2$ I believe this can still be made to work: Whitney trick to make $f$ an embedding and also some care about stable vs unstable normal bundle.
The smallest dimension is 2 : you may take the open annulus and Möbius band : their tangent bundles are both diffeomorphic to $S^1\times R^3$. I think you can obtain $k=\infty$ in three dimensions, by taking various Whitehead manifolds (contractible 3-manifolds not diffeomorphic to $R^3$ but whose product with $R$ is diffeomorphic to $R^4$, if I'm not mistaking, see this question, answers and comments).
For $k = \infty$ (a continuum, to be precise), the continuum of non-diffeomorphic smooth structures on $\mathbb R^4$ would suffice. The tangent bundle of any $\mathbb R^4$ is trivial (since $\mathbb R^4$ is contractible), therefore homeomorphic to $\mathbb R^8$, but $\mathbb R^8$ has only one smooth structure up to diffeomorphism.
