$\DeclareMathOperator\GL{GL}\DeclareMathOperator\L{\mathfrak{L}}$The free Lie algebra $\L(V)$ generated by an $r$-dimensional vector space $V$ is, in the
language of https://en.wikipedia.org/wiki/Free_Lie_algebra,
the free Lie algebra generated by any choice of basis $e_1, \ldots , e_r$ for the vector space $V$. (Work over the field ${\mathbb R}$ or ${\mathbb C}$, whichever you prefer.)
It is a graded Lie algebra

$$\L(V) = V \oplus \L_2 (V) \oplus \L_3 (V) \oplus \ldots .$$
The general linear group $\GL(V)$ of $V$ acts on $\L(V)$ by gradation-preserving Lie algebra automorphisms.
Thus each graded piece $\L_k (V)$ is a finite dimensional
representation space for $\GL(V)$. (The `weight' of $\L_k (V)$ is $k$ in the sense that $\lambda \mathrm{Id} \in \GL(V)$ acts on $\L_k (V)$ by scalar multiplication by $\lambda^k$.)
QUESTION: How does $\L_k (V)$ break up into $\GL(V)$-irreducibles?

I only really know that $\L_2 (V) = \Lambda ^2 (V)$, which is already irreducible.

To start the game off, perhaps some reader out there already is familiar with $\L_3 (V)$ as a $\GL(V)$-rep, and can tell me its irreps in terms of the Young diagrams / Schur theory involving 3 symbols?

(My motivation arises from trying to understand some details of the subRiemannian geometry https://en.wikipedia.org/wiki/Sub-Riemannian_manifold of the Carnot group whose Lie algebra is the free $k$-step Lie algebra, which is $\L(V)$-truncated after step $k$. )