There is no such thing as "the shortest program for which the halting property is uncomputable". That property is computable trivially for every possible program.

The halting property is semi-decidable: if a program halts, then you can always trivially prove that it halts, you just need run it. If a program does not halt, then sometimes you can prove that it doesn't halt, and sometimes you can't prove anything.
For any programming language, there exist a length such that you can compute the halting property for all programs shorter than that. At length L_max, there will be program Bad_0, which:

• Does not halt.
• Can't be proven not to halt.
• Doesn't emit any output symbol.
• It can't be proven that there exist any string of output symbols that it doesn't emit.

You can never prove that any string S has Kolmogorov complexity K if K > L_max, as it would imply that you proved that Bad_0 doesn't halt, or at least doesn't emit any symbol which is not a prefix of S.

Since there are only finitely many strings with complexity up to L_max, we can only compute Kolmogorov complexity, for finitely many strings.

It is also easy to make up artificial languages in which Kolmogorov complexity is computable for an infinite subset of all possible strings.

If the language is Turing-complete I don't think this is possible. If you think that an arbitrary string S has complexity K, how can you prove that there exist no program shorter than K that computes S?