Kolmogorov Complexity of a Decidable Language
Yes, depending on what kinds of inputs you consider (see below). $KC(x) =^* KCDL(L_x)$, where $L_x$ is the language which consists only of the string $x$, and $=^*$ means equals up to an additive constant.

The reverse is probably not possible (I think I proved this at one point but can't find it right now). The idea is that Kolmogorov complexity can be solved with a Halting Problem oracle, but KCDL (when the input is a TM) is, if I recall correctly, hard for a higher level of the arithmetic hierarchy.

If you want to talk about the complexity of KCDL you also have to be careful to define what the input is. You want the input to be some finite object - languages are generally not. One natural thing for decidable languages is that the input should be a (total) TM. Another possibility is that the input is a TM, along with a proof (in some fixed, formal language) that the TM is total. Other possibilities exist too.

I haven't seen published work about it, but I thought about it a bit in grad school and proved/observed some elementary results. You can do the same for any computable function, not just a language.