In my opinion, it's probably not so interesting to consider all extensions. If you consider the Lie subalgebra $\mathfrak{n}_n\subset \mathfrak{gl}_n$ of all strictly upper triangular matrices, then $\mathfrak{n}_n$ is an extension of $\mathbf{C}$ because it has a quotient Lie algebra of dimension 1. It follows that even for nilpotent extensions the nilpotency can grow arbitrarily large. Of course extensions can be even more chaotic than that.

In my opinion, it's probably not so interesting to consider all extensions. If you consider the Lie subalgebra $\mathfrak{n}_n\subset \mathfrak{gl}_n$ of all strictly upper triangular matrices, then $\mathfrak{n}_n$ is an extension of $\mathbf{C}$ because it has a quotient Lie algebra of dimension 1. It follows that even for nilpotent extensions the nilpotency can grow arbitrarily large. Of course extensions can be even more chaotic than that.

Edit Example of a non-central perfect group extension: Consider block matrices over $\mathrm{C}$ $$ H = \begin{pmatrix}\mathrm{SL_2} & * \\ 0 & \mathrm{SL_2}\end{pmatrix},\; K = \begin{pmatrix}I_2 & * \\ 0 & I_2\end{pmatrix},\; G = H/K \cong \mathrm{SL}_2\times \mathrm{SL}_2 $$ The group $H$ is perfect because its Lie algebra is perfect: The diagonal part $\mathfrak{sl}_2\times \mathfrak{sl}_2$ lies in the commutator because $\mathfrak{sl}_2$ is perfect. On the other hand, we have $$ X = \begin{pmatrix}0 & 0 & b & a \\ 0 & 0 & d & c \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}, Y = \begin{pmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix},\; [X, Y] = \begin{pmatrix}0 & 0 & a & b \\ 0 & 0 & c & d \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix} $$ so the off-diagonal part is also in the commutator.

Consider the conjugation by the matrix $\mathrm{diag}(1, 1, -1, -1)$. It acts non-trivially on $H$ but trivially on $G$. It means that non-central extensions can in general have non-trivial automorphisms even if the extension is a perfect group.