To which extent can one recover a manifold from its group of homeomorphisms

Answer is: Yes, one can recover $M$ if it is a compact manifold. See J. V. Whittaker: On Isomorphic groups and homeomorphic spaces, Annals of Math 1963.

EDIT Actually, one knows a lot more, see, for example Tomasz Rybicki Journal: Proc. Amer. Math. Soc. 123 (1995), 303-310. MSC: Primary 58D05; Secondary 17B66, 22E65, 57R50 MathSciNet review: 1233982

And references therein...

ANOTHER EDIT

A quite different proof of a stronger theorem (actually a large set of theorems) than Whittaker's (actually, Whittaker's paper seems to be rather badly written) is given by Matatyahu Rubin in Rubin, Matatyahu(3-SFR) On the reconstruction of topological spaces from their groups of homeomorphisms. Trans. Amer. Math. Soc. 312 (1989), no. 2, 487–538.

This is more of a longish comment rather but I'd like to point out that while Igor's reference in principle gives a complete answer, actually reading off any specific information about $M$ (such as its dimension) from some topological invariants of $Diff(M)$ is likely hard.

Moreover, if one relaxes the categories somewhat then the answer to Misha's question can even be negative! Specifically, one can ask if the homotopy type of the monoid of self homotopy equivalences of $M$ determines $M$ up to homotopy type. I actually don't know the answer to this but if one relaxes the category even further and looks at the rational homotopy type then in contrast with the diffeomorphism case the answer is actually NO.

Specifically, it's rather easy to compute that the rational homotopy type of the identity component $Aut(M)$ of the monoid of self homotopy equivalences of an equal rank biquotient of Lie groups $M=G//H$ that satisfies Halperin's conjecture (which says that in this case $H^\ast (M,\mathbb Q)$ has no negative degree derivations) is determined by rational homotopy and homology groups of $M$. In this case $Aut(M)$ is rationally equivalent to a product of finitely many odd dimensional spheres and one can write an explicit (if somewhat ugly) formula for the dimensions of the spheres that show up in terms of $\pi_\ast(M)\otimes \mathbb Q$ and $H_\ast(M,\mathbb Q)$.

But there are plenty of examples of such biquotients in dimensions above 5 which have distinct rational types but the same rational homotopy and homology. For example, one can take $G//T$ where $G$ is a simply connected Lie group and $T\le G\times G$ is a torus of the same rank as rank $G$. All such biquotients satisfy Halperin's conjecture so the formula I mention above applies. It is then clear that rational homology and homotopy groups of $G//T$ are completely determined by $G$ but the rational type of $G//T$ can be different depending on the embedding $T\to G\times G$. There are infinitely many such examples already in dimension 6 of the form $(S^3\times S^3\times S^3)//T^3$. Still, in this case one can read off for example, the dimension of $M$ from the knowledge of the rational homotopy groups of $Aut(M)$ but I don't know how to get such formula for a general closed simply connected manifold $M$ (and I'm not even sure if it's possible).

For the smooth case, the result is in:

• Takens, F. (1979). Characterization of a differentiable structure by its group of diffeomorphisms. Bol. Soc. Brasil. Mat., 10, 17–25. MR552032

and the answer is "Yes".

For completeness, Takens' theorem is:

Theorem Let $\Phi \colon M_1 \to M_2$ be a bijection between two smooth $n$-manifolds such that $\lambda \colon M_2 \to M_2$ is a diffeomorphism iff $\Phi^{-1} \circ \lambda \circ \Phi$ is a diffeomorphism. Then $\Phi$ is a diffeomorphism.