Every known construction of the Vitali set relies on the Axiom of Choice. It happens to not be Lebesgue-measurable.
Must every set whose construction relies on the Axiom of Choice not be Lebesgue-measurable?
I'm not sure if "relies on the Axiom of Choice" is mathematically meaningful, in which case I'll rephrase the question as asking whether there are any sets for which there is an existence proof with AC but no known existence proof without AC, which are known to be Lebesgue-measurable.
This related question asks whether there exists a Lebesgue non-measurable set that can be constructed without AC. The answer to that question is no.