- Indeed, $MCG(\mathbb T^n)=GL(n,\mathbb Z)$ in dimension $n<4$, but it is not simple. In dimension 2 it was first proved by Earle and Eells using complex analysis.[Edit: As Allen Hatcher points out this was known for a long time, Earle and Eells prove much stronger statement: $\mathbb T^2$ is deformation retraction of $Diff_0(\mathbb T^2)$]
- I am not sure what happens in dimension 4.
- This is not correct in dimension >4, the MCG is semidirect product of $GL(n,\mathbb Z)$ with another (non-finitely generated group). Let me just quote Hatcher:
If $n\ge 5$ then there are split exact sequences
$$
0\to \mathbb Z_2^\infty\to\pi_0(Top(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
$$
$$
0\to \mathbb Z_2^\infty\oplus\binom n2\mathbb Z_2\to\pi_0(PL(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
$$
$$
0\to \mathbb Z_2^\infty\oplus\binom n2\mathbb Z_2\oplus\sum_{i=0}^n\binom n i\Gamma_{i+1}\to\pi_0(Diff(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
$$
where $\Gamma_i$ are Kervaire-Milnor finite abelian groups of homotopy spheres, $\mathbb Z_2$ is just the group of order 2 and $\mathbb Z_2^\infty$ is the group of finite strings.
The above quote is from Hatcher "Concordance spaces, higher simple homotopy theory and application."