Are the mapping class groups of manifolds finitely presentable?

These things can be pretty wild. For instance, for $n \geq 5$ the mapping class group of the $n$-torus is not even finitely generated : it is a split extension of $\text{GL}_n(\mathbb{Z})$ by an infinite rank abelian group. See Theorem 4.1 of

A. E. Hatcher, Concordance spaces, higher simple-homotopy theory, and applications. Proc. Sympos. Pure Math., XXXII, Part 1, pp. 3-21, Amer. Math. Soc., Providence, R.I., 1978.

and Example 4 of

Hsiang, W. C.; Sharpe, R. W. Parametrized surgery and isotopy. Pacific J. Math. 67 (1976), no. 2, 401–459.

I also recommend Hatcher's survey about the homotopy type of the diffeomorphism groups of manifolds here; it discusses a lot of things that are known about $\pi_0$.

It should be remarked that the above trouble happens because we are considering manifolds with infinite fundamental groups, which often causes terrible things to happen in high-dimensional topology. If $M$ is a compact simply-connected manifold of dimension at least $5$, then Sullivan proved that the mapping class group of $M$ is an arithmetic group, and hence has every possible finiteness property. This was extended to manifolds with finite fundamental groups by Triantafillou. For all of this, I recommend the survey

Triantafillou, Georgia The arithmeticity of groups of automorphisms of spaces. Tel Aviv Topology Conference: Rothenberg Festschrift (1998), 283–306, Contemp. Math., 231, Amer. Math. Soc., Providence, RI, 1999.

The mapping class group $\pi_0 Diff(S^1 \times D^3)$ is not finitely generated.

The proof of this amounts to arguing that $Diff(S^1 \times D^3)$ acts transitively on the reducing discs, i.e. the properly-embedded $3$-discs in $S^1 \times D^3$ that are non-separating. The isotopy classes of reducing discs form a group, and that is not finitely-generated.

We show this by arguing that any component of the space of embeddings $S^1 \to S^1 \times S^3$ has a not-finitely-generated fundamental group. This part of the argument is fairly classical: Haefliger-style double-point considerations. This step appears to have been known to Dax, as well as Arone and Szymik.

Then we look at the fibre bundle $Diff(S^1 \times S^3) \to Emb(S^1, S^1 \times S^3)$. The fiber over the "degree 1 component" of the embedding space is the diffeomorphisms of $S^1 \times S^3$ that fix the knot pointwise, which is closely related to $Diff(S^1 \times D^3)$. The key argument is showing that none of our "interesting" elements of $\pi_1 Emb(S^1, S^1 \times S^3)$ are in the image of the map from $\pi_1 Diff(S^1 \times S^3)$. By an isotopy-extension argument they are turned into non-trivial isotopy classes of diffeomorphisms of $S^1 \times D^3$.