Let $L$ be a sub-algebra of $\mathrm{Vect}(M)$. I think one might be able to prove $\mathrm{codim}\ L \geq \dim M$ by using a recent result of Hurtado. Here is a sketch of the proposed proof, every step of which is difficult:

(1) Let $G \subset \mathrm{Diff}(M)$ be the group generated by $\exp(L)$. **Problem:** The exponential need not converge.

(2) Let $N = \mathrm{Diff}(M)/G$. One hopes that $N$ is a manifold of dimension $\mathrm{codim}(L)$. **Problem** It is not clear how to put a manifold structure on $N$.

We then get a homomorphism $\mathrm{Diff}(M) \to \mathrm{Diff}(N)$ by the action of $\mathrm{Diff}(M)$ on the quotient $N$. Hurtado shows that such maps only exist if $\dim M \leq \dim N$.

We can, at least momentarily, dodge discussions of quotients and exponentials of infinite dimensional Lie groups by defining $N$ to be be the set of subalgebras of $\mathrm{Vect}(M)$ conjugate to $L$ by an element of $\mathrm{Diff}(M)$ (or perhaps $\mathrm{Diff}_0(M)$.)

Hopefully, the set of all codimension $n$ subspaces of $\mathrm{Vect}(M)$ is some sort of manifold by a Grassmannian like construction, and then $N$ can be a closed submanifold of this. If we are lucky, then the tangent space to $N$ at $[L]$ is $\mathrm{Vect}(M)/L$.

We clearly have a map $\mathrm{Diff}(M) \to \mathrm{Bijections}(N)$ and, presumably, once we have built the smooth structure on $N$ this will be a map $\mathrm{Diff}(M) \to \mathrm{Diff}(N)$ and Hurtado's result will tell us that $\dim M \leq \dim N = \dim(\mathrm{Vect}(M)/L)$.

All of this needs to come with the disclaimer that I am a simple minded algebraic geometer, who doesn't like these infinite dimensional objects.