Let $\mathfrak s$ be a complex semisimple Lie algebra, then $\dim _\mathbb C\mathfrak s\geq 3$. But, however, is it possible for $\mathfrak s$ to have a semisimple Lie subalgebra $\mathfrak h$ such that $\dim_\mathbb C\mathfrak s/\mathfrak h=2$?
On Lie groups level, do homogeneous spaces of the form $\mathrm{SL}(n,\mathbb C)/H$ where $H$ is a closed complex semisimple subgroup of $\mathrm{SL}(n,\mathbb C)$ have special properties, for example parallelizable, etc?
No, it's not. Indeed it's known that no simple Lie algebra of rang $\ge 3$ has a proper subalgebra of codimension 2; in rank 2, the only proper subalgebras of codimension $\le 2$ are parabolic of codimension 2, and in rank 1 the only proper subalgebras of codimension $\le 2$ are solvable (either parabolic or abelian).
So if $\mathfrak{g}$ is a semisimple Lie algebra and $\mathfrak{h}$ is a proper subalgebra of codimension $\le 2$ then there exists a direct decomposition $\mathfrak{g}=\mathfrak{g}_1\times\mathfrak{g}_2$ and $\mathfrak{h}=\mathfrak{g}_1\times\mathfrak{h}_2$ with either $\mathfrak{h}_2$ parabolic in $\mathfrak{g}_2$ (hence not semisimple), or $\mathfrak{g}_2$ has dimension 3 and $\mathfrak{h}_2$ is abelian (hence not semisimple).
Codimension 3 is possible, namely in $\mathfrak{sl}_2^2$.