This is an attempt to prove the (refined) conjecture I made in the comments of my previous answer. Let $\mathfrak{g}$ be a f.d soluble Lie algebra over $\mathbb{C}$. Let $\mathfrak{n}$ be its derived subalgebra which is nilpotent by Lie's theorem. I claim that for $\lambda\in ( \mathfrak{g}/\mathfrak{n})^\ast$, $H_i(\mathfrak{g}, \mathbb{C}_\lambda)\neq 0$ only if $\lambda$ is a sum of $i$ weights in the adjoint action.
To see this we pick a one-dimensional ideal $\mathfrak{z}=\mathbb{C}z$ in $ \mathfrak{g}$ contained in $\mathfrak{n}$. [If $\mathfrak{n}=0$ the result is easy by the now commutative methods in my previous answer.] Let's say z has weight $\mu$.
The Hochschild-Serre spectral sequence says that $H_i(\mathfrak{g}/\mathfrak{z} , H_j(\mathfrak{z} ,\mathbb{C}_\lambda))$ converges to $H_{i+j}(\mathfrak{g},\mathbb{C}_\lambda)$.
So we first understand $H_j(\mathfrak{z},\mathbb{C}_\lambda)$ as a $\mathfrak{g}$-module. This can be computed as $\mathrm{Tor}_j^{U(\mathfrak{g})}(U(\mathfrak{g})/(z), \mathbb{C}_\lambda)$ which is the $j$th homology group of $$0\to U(\mathfrak{g})\otimes_{U(\mathfrak{g})} \mathbb{C}_\lambda \to U(\mathfrak{g})\otimes_{U(\mathfrak{g})} \mathbb{C}_\lambda\to 0$$ where the non-zero map is multiplication by $z$. It follows that $$H_0(\mathfrak{z},\mathbb{C}_\lambda)=\mathbb{C}_\lambda$$ and $$H_1(\mathfrak{z}, \mathbb{C}_\lambda)=\mathbb{C}_{\lambda-\mu}.$$ By induction on $\dim\mathfrak{g}$ we're done.