The answers to your questions are 'no' and 'yes'.

In the first place, there is no finite dimensional manifold whose Lie algebra of vector fields contains all of the Lie algebras ${\frak{sl}}(n,\mathbb{R})$ for all $n\ge 2$.

In fact, ${\frak{sl}}(n{+}2,\mathbb{R})$ cannot occur as a Lie algebra of vector fields on any $n$-manifold. The reason is that this Lie algebra does not have any nontrivial subalgebras of codimension less than $n{+}1$. If $\frak{g}$ were a subalgebra of the vector fields on $M^n$, then, for any $x\in M$, the subalgebra ${\frak{g}}_x$ consisting of the elements of $\frak{g}$ that vanish at $x$ would have codimension at most $n$ and hence would have to satisfy ${\frak{g}}_x={\frak{g}}$, i.e., all of the vector fields in ${\frak{g}}$ would have to vanish at all of the points of $M$, which is absurd.

In the second place, $M = \mathbb{R}^2$ has the property that there is a Lie subalgebra of dimension $n$ of the vector fields on $M$ for every $n\ge0$.

Technically, you can have Lie subalgebras of the vector fields on $\mathbb{R}^1$ of arbitrarily high dimension: Just let $X_k$ be any vector field on $\mathbb{R}^1$ with compact support contained in the interval $[k,k{+}1]$. Then the $X_k$ all Lie-commute and the algebra that they generate contains the $n$-dimensional (abelian) subalgebra spanned by $X_1,\ldots, X_n$.

However, this is kind of a cheating example since, at any one point, you can only 'see' a $1$-dimensional subalgebra. What you'd really like is an example in which the entire algebra injects into the vector fields on any open subset when you restrict to that open subset. With that extra assumption, you can't have finite dimensional Lie algebras of arbitrarily high dimension acting on $\mathbb{R}^1$; you have to go to dimension $2$ at least.

In dimension $2$, you can have arbitrarily high dimensional Lie subalgebras of, say the torus: Let $x$ and $y$ be $2\pi$-periodic coordinates on the torus, and consider the vector fields
$$
X = \frac{\partial\ }{\partial x}
\qquad\text{and}\qquad
Y_k = \cos kx\ \frac{\partial\ }{\partial y}
\qquad\text{and}\qquad
Z_k = \sin kx\ \frac{\partial\ }{\partial y}.
$$
Then, for any $n$, the vector fields $X,Y_0,\ldots,Y_n,Z_0,\ldots,Z_n$ form a (solvable) Lie algebra of vector fields on the torus. Moreover, they are the infinitesimal generators of a transitive Lie group action on the torus.