Your title question and your body are different: your title asks what Lie algebras arise as fcss centralisers, and your body asks whether all Lie algebras arise this way.  I answer the easier latter question; I have no idea what shape an answer to the former would take, aside from that I can produce a countable list of isomorphism types by using the classifications of root systems and of nilpotent orbits as below, and ask you to check whether your Lie algebra is isomorphic to one on that list.

Specifically, I claim first that there are only countably many fcss centralisers.  It clearly suffices to show the same with ‘reductive’ in place of ‘semisimple’; I guess we can call that ‘fcr’.  First note that, if $\mathfrak g$ is reductive and $x$ is an element of $\mathfrak g$ with Jordan decomposition $x = x_\text s + x_\text n$, then $\mathfrak m \mathrel{:=} \mathfrak g^{x_\text s}$ is reductive, and $\mathfrak g^x$ equals $\mathfrak m^{x_\text n}$.  It thus suffices to show that there are only countably many isomorphism types of centralisers of nilpotent elements in an fcr Lie algebra.  First note that, since an fcr Lie algebra is classified by the dimension of its centre and the root datum of its derived subalgebra, there are only countably many fcr Lie algebras.  Since each fcr Lie algebra has only finitely many nilpotent orbits under the associated adjoint group, we have shown the claim.

@YCor's [answer](https://mathoverflow.net/a/165697) to https://mathoverflow.net/questions/165656/how-many-three-dimensional-real-lie-algebras-are-there shows that there are uncountably many isomorphism types of 3-dimensional complex Lie algebras.  Thus, uncountably many of them are not fcss centralisers.