Suppose that $G$ is a complex semisimple Lie group, $P$ a parabolic subgroup of $G$. What are all of the $P$invariant subspaces of $\mathfrak{g}/\mathfrak{p}$? In various low dimensional examples, I can calculate them all out, but there should be some easy way to describe each of them in terms of the Dynkin diagram of $G/P$.

While I share the doubts of others about finding a magic recipe, I'll add some further remarks about the formulation of the problem, which has already been mostly translated into the language of Lie algebras. Fix a semisimple complex Lie algebra $\mathfrak{g}$ plus a (proper) subset $I$ of the simple roots relative to a Cartan subalgebra $\mathfrak{h}$ and a Borel subalgebra $\mathfrak{b}$ containing it, together with a standard parabolic subalgebra $\mathfrak{p}_I = \mathfrak{l}_I \oplus \mathfrak{u}_I$. Here $\mathfrak{l}_I$ is a reductive Levi subalgebra with simple roots $I$ and $\mathfrak{u}_I$ is spanned by 1dimensional root spaces belonging to positive roots not in the span of $I$. Now the Lie algebra problem is to determine all $\mathfrak{p}_I$stable subspaces of $\mathfrak{u'}_I$, the opposite nilradical. These are sums of negative root subspaces closed under addition of positive roots not in the span of $I$ and also closed under the action of $\mathfrak{l}_I$, hence direct sums of various irreducible modules for the semisimple derived algebra of $\mathfrak{l}_I$. 1) In case $I$ is empty (so $\mathfrak{p}_I = \mathfrak{b}$), such a subspace is a sum of the easiest types of submodules: a single negative root space (say for $\beta<0$) along with all others obtained by adding positive roots to $\beta$. But listing or counting these seems to require brute force enumeration, using data on positive roots for each simple Lie algebra. See the planches at the end of Bourbaki, Lie Groups and Lie Algebras, Chap. IVVI, or for exceptional types the more convenient explicit lists (ordered by height) at the end of Springer's old paper here. A look at the higher rank exceptional types shows how unpredictable the end results will be. 2) For a larger parabolic $\mathfrak{p}_I$, you also need to keep track of the irreducible modules for $\mathfrak{l}_I$ which occur. This gets complicated, especially if there are numerous simple Lie algebra summands in this reductive Lie algebra. Even for type $G_2$, the Levi subalgebra corresponding to a short simple root requires a little care even though only two irreducible summands of the 5dimensional opposite nilradical will exist (and their sum is stable under the full parabolic). But in general there are $2^\ell$ nonconjugate types of parabolics if the rank is $\ell$. Lots of work. 


A $P$invariant subspace is $T$invariant, hence it is a sum of weight spaces. So, each such subspace corresponds to a subset of the set of positive roots. The additional condition of $U$invariance forces these subsets to have the following property: if $\alpha$ is in subset and $\beta  \alpha$ is a sum of positive roots then $\beta$ is in the subset. EDIt: All this is for the case $P = B$. For arbitrary parabolic the situation is more complicated, see robot's answer. 


This is not an answer. Just a few well known facts. Each $P$invariant subset is also invariant with respect to the Levi part $L$ of $P$ and hence it decomposes into irreducibles for $L$. The representation $\mathfrak{g/p}$ is as a $\mathfrak{p}$representation isomorphic (via the Killing form) to the nilradical of $\mathfrak{p}$. Now this nilradical is in fact isomorphic to $k$graded Lie algebra $\bigoplus_{i=1}^k \mathfrak{g}_i$ that is generated (as a Lie algebra) by $\mathfrak{g}_1$. Lie brackets $\mathfrak{g}_i \otimes \mathfrak{g}_j \to \mathfrak{g}_{i+j}$ for $i,j\in \{1,\ldots k\}$ are $L$equivariant. (See e.g. section 3.1.2 in Parabolic Geometries I Background and General Theory by Čap and Slovák.) One always have $P$invariant subspaces given by filtration components $\mathfrak{g}^j = \bigoplus_{i=j}^k \mathfrak{g}_i$. Consider a $P$invariant subspace $V$. If $\emptyset \neq (V\cap \mathfrak{g}_i) \neq \mathfrak{g}_i$, then one can perhaps use the generating property of $\mathfrak{g}_1$ and $L$equivariance of the Lie bracket $\mathfrak{g}_1\otimes (V\cap \mathfrak{g}_i) \to \mathfrak{g}_{i+1}$ to restrict possible $L$types occurring in $V$. But I don't know how the $L$decompositions of $\mathfrak{g}_i$ look like. 

