I am interested to identify (ideally classify) nilpotent Lie algebras that occur as nilradicals of parabolic subalgebras in (say) reductive Lie algebras. For example, all Heisenberg Lie algebras appear as such, the same holds for free 2-step nilpotent Lie algebras. But what about general free n-step nilpotent Lie algebras?

Let $\mathfrak{m}_n$ denote the free 2-step nilpotent Lie algebra with $n$ generators over $\mathbb{C}$. Consider the simple Lie algebra $\mathfrak{g} = \mathfrak{so}(2n+1)$ and the set of positive roots $$\Delta_+ = \{e_i - e_j, e_i + e_j \mid 1 \leq i \lt j \leq n\} \cup \{e_i \mid 1 \leq i\leq n\}.$$ So we have
$$\mathfrak{g} = \mathfrak{n}_- \oplus \mathfrak{gl}(n) \oplus \mathfrak{n}_+$$
whith a root space decomposition of
$$\mathfrak{n}_+ = \bigoplus_{i} \mathfrak{g}_{e_i} \oplus \bigoplus_{i \lt j} \mathfrak{g}_{e_i + e_j}$$

We see that $\mathfrak{m}_n \simeq \mathfrak{n}_+ $ appears as a nilradical of the parabolic $\mathfrak{p} = \mathfrak{gl}(n) \oplus \mathfrak{n}_+ $.
For a free nilpotent Lie algebra of higher index (say 3), we'd need to find a (perhaps generalized) root system that "matches" the corresponding Hall basis. The reductive part $\mathfrak{gl}(n)$ would stay the same as long as we fix $n$ as the number of generators.

The dim$(\mathfrak{g}^{k}) \gt d^2$ argument in Yves' answer below is indeed strong. Still, I was wondering if one can start to look at this from the other way around. I.e. starting with any finite-dimensional nilpotent Lie algenbra $\mathfrak{n}$ over $\mathbb{C}$, then looking at the derivation algebra Der$(\mathfrak{n})$ of $\mathfrak{n}$, deducting the reductive part $\mathfrak{l}$ of Der$(\mathfrak{n})$ (let's assume that $\mathfrak{n}$ is not characteristically nilpotent) so $\mathfrak{l}$ acts on $\mathfrak{n}$ in a natuarl way. We may then try to treat $\mathfrak{n}$ similar as it was a nilpotent radical of something like $\mathfrak{l} \oplus \mathfrak{n}$ and hence constucting a Lie algebra like $$\mathfrak{g} := \mathfrak{n}_- \oplus \mathfrak{l} \oplus \mathfrak{n}$$ where $\mathfrak{n}_-$ is another copy of $\mathfrak{n}$ analogous to a root space decomposition. This works particularly fine with the free nilpotent Lie algebras $\mathfrak{f}(k,d)$ where in this case we have $\mathfrak{l} = \mathfrak{gl}_d$. I was thinking if it would be possible to define something like roots and a Weyl group to the above constructed $\mathfrak{g}$. Perhaps in the sense of G. Favre, Système de poids sur une algèbre de Lie nilpotente, Manuscripta Math. 9, 1973. However, I'd love to see this in the context of the bigger $\mathfrak{g}$ using the "big" Weyl group $W(\mathfrak{g})$ as well as the "small" Weyl group $W(\mathfrak{l})$. So I was just wondering if at all it makes sense to glue together this $\mathfrak{g}$ as described above. It will in general surely not be semi-simple anymore as we have found out, but perhaps something "similar".

allfree 2-step nilpotent as nilradical of parabolics? $\endgroup$ – YCor Mar 3 '14 at 21:54