Modules over the unipotent subalgebra as direct summands of modules over a semisimple Lie algebra

Question

Let $\mathfrak g$ be a semisimple finite-dimensional Lie algebra over the field of complex numbers $\mathbb C$. Let $\mathfrak n\subset\mathfrak g$ be the maximal unipotent subalgebra of $\mathfrak g$. (E.g., for the algebra of all traceless $n\times n$ matrices $\mathfrak g=\mathfrak{sl}(n,\mathbb C)$, the subalgebra $\mathfrak n$ is the Lie algebra of all strictly upper-triangular $n\times n$ matrices, with zeroes on the diagonal.)

Let $N$ be a finite-dimensional nilpotent $\mathfrak n$-module (i.e., an $\mathfrak n$-module in which all the elements of $\mathfrak n$ act by nilpotent linear operators). Can I find a finite-dimensional $\mathfrak g$-module $M$ such that $N$ is a direct summand of the underlying $\mathfrak n$-module of $M$? If not, can I find an infinite-dimensional $\mathfrak g$-module $M$ with the similar property (generally speaking)?

For $\mathfrak g=\mathfrak{sl}(2,\mathbb C)$, the answer is positive. In this case, $\mathfrak n$ is a one-dimensional abelian Lie algebra with a generator $x$, and finite-dimensional nilpotent $\mathfrak n$-modules are finite direct sums of the $k[x]$-modules $k[x]/(x^m)$, where $m\ge0$ are nonnegative integers. The $\mathfrak n$-module $k[x]/(x^m)$ underlies the (unique) $m$-dimensional irreducible $\mathfrak g$-module. So, in this case, every finite-dimensional nilpotent $\mathfrak n$-module admits an extension of its $\mathfrak n$-module structure to a $\mathfrak g$-module structure, and one can even have $M=N$.

Is the answer negative for semisimple Lie algebras $\mathfrak g$ of higher rank?

(My motivation is far away from the Lie algebra representation theory. For reasons having to do with the theory(ies) of corings and bocses, I am looking for an associative ring $A$ with a subring $B$ having the following properties:

1. the ring $A$ has finite left global (homological) dimension;
2. the ring $A$ is a free (or faithfully projective, or at least faithfully flat) right $B$-module, preferably also a free (or faithfully projective) left $B$-module;
3. the full subcategory of all $B$-module direct summands of underlying $B$-modules of left $A$-modules is not closed under extensions in the abelian category $B\mathrm{-Mod}$.

It is easy to satisfy 1. and 2., not difficult to satisfy 1. and 3., and I know how to satisfy 2. and 3. Looking for a way to satisfy all the three conditions, I came up with the idea of considering $A=U(\mathfrak g)$ and $B=U(\mathfrak n)$. End of motivation.)