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". 



Which nilpotent Lie algebras appear as nilradicals of parabolic subalgabras?
Among free nilpotent Lie algebras the possible nilradicals of parabolics are exactly (up to 1 case):


* the abelian ones
* the 2-nilpotent ones (as you mentioned)
* the free 3-nilpotent on 2 generators (5-dimensional).


The latter appears as nilradical of a 9-dimensional parabolic subalgebra of the (14-dimensional) exceptional simple Lie algebra of type $G\_2$. 


It remains to exclude the other ones. One way to do so (which also works for many other nilpotent Lie algebras) is to use the following fact: if $P$ is a parabolic subgroup in a semisimple complex algebraic group with unipotent radical $P\_u$, and $A$ is a normal abelian subgroup of $P$ contained in $P\_u$, then $P$ has finitely many orbits on $A$. This result is due to *G. Röhrle, On normal abelian subgroups in parabolic groups. Ann. Fourier 48(5) (1998) 1455-1482;* still I'll only use it in the case of a subgroup central in $P\_u$, which Röhrle claims as a standard well-known case (as well as the characteristic zero case, which he attributes to earlier work).


It remains to deduce the above list. Let $\mathfrak{g}$ be a nilpotent Lie algebra of nilpotency length $k$ and whose abelianization $\mathfrak{a}$ has dimension $d$. Let $\mathfrak{g}^k$ be the last term in its lower central series. Then the automorphism group of $\mathfrak{g}$ acts on $\mathfrak{g}^k$, in such a way that those automorphisms acting trivially on $\mathfrak{a}$ also act trivially on $\mathfrak{g}^k$. Hence the action of $\mathrm{Aut}(\mathfrak{g})$ on $\mathfrak{g}^k$ factors through a subgroup of $\mathrm{GL}\_d$. Therefore, if $\dim(\mathfrak{g}^k)>d^2$ then the action of $\mathrm{Aut}(\mathfrak{g})$ on $\mathfrak{g}^k$ has infinitely many orbits.


It remains to compute this in the case of the free $k$-nilpotent Lie algebra $\mathfrak{f}(k,d)$ on $d$ generators. For $k=3$, we have $\dim(\mathfrak{f}(3,d)^3)=(d^3-d)/3$, which is $>d^2$ as soon as $d\ge 4$ (for $(k,d)=(3,3)$ see below). For $k=4$, we have $\dim(\mathfrak{f}(4,d)^4)=(d^4-d^2)/4$, which is $>d^2$ as soon as $d\ge 3$ (for $(k,d)=(4,2)$ see below). For $k=5$, we have $\dim(\mathfrak{f}(5,d)^5)=(d^5-d)/5$ which is $>d^2$ for all $d\ge 2$, since $\dim(\mathfrak{f}(k,d)^k)$ increases, for $d$ fixed, when $k\ge 2$, this rules out $k\ge 5$ as well.


It remains $(k,d)=(3,3)$ or $(4,2)$. In the case $(4,2)$, the dimension of $\dim(\mathfrak{f}(4,2)^4)$ is 3, so the Levi part of $P$ should be at least 3 (while it's at most 4) since it embeds into $\mathrm{GL}\_2$). The dimension of $\dim(\mathfrak{f}(4,2))$ is $2+1+2+3=8$, so the dimension of the Lie algebra should be $2\times 8+3/\!\!/4=19/\!\!/20$, but there is no simple complex Lie algebra of this dimension (and non-simple semisimple are excluded, in view of the comments). Hence $(4,2)$ is discarded. In the case $(3,3)$, the dimension of $\dim(\mathfrak{f}(3,3)^3)$ is 8, so the Levi part of $P$ should be at least 8 (while it's at most 9 since it embeds into $\mathrm{GL}\_3$). The dimension of $\dim(\mathfrak{f}(3,3))$ is $3+3+8=14$. Hence the dimension of the Lie algebra should be $2\times 14+8/\!\!/9=36/\!\!/37$ (// means or). By the classification of simple complex Lie algebras, there are 2 possibilities then, namely $\mathrm{SO}\_9$ and $\mathrm{Sp}\_8$, both 36-dimensional. But here it means that the Levi subalgebra is $\mathrm{SL}\_3$, and it follows (by a simple argument) that $P$ is unimodular, which is a contradiction.


