The instruction set doesn't matter. To calculate (the first few bits of) Chaitin's Constant we can enumerate and run every program in some instruction set, say x86 assembly, and measure the proportion of programs which eventually halt. One efficient way to do this is to interleave their executions using FAST http://www.idsia.ch/~juergen/toesv2/node28.html
Any (Turing-complete) instruction set can be translated/compiled/interpreted-by any other (Turing-complete) instruction set, given a suitable translator/compiler/interpreter. Since we're running all programs, we will eventually run all translators/compilers/interpreters, so no matter what instruction set we choose, we will start running programs from every other. The longer we leave it running, the more of a mixture we end up with. Since Chaitin's Constant is the (uncomputable) result of letting such a scheme run forever, it contains a perfect mixture of all instruction sets, and is thus independent of whichever one we choose.
Any (Turing-complete) instruction set can be translated/compiled/interpreted-by any other (Turing-complete) instruction set, given a suitable translator/compiler/interpreter. Since we're running all programs, we will eventually run all translators/compilers/interpreters, so no matter what instruction set we choose, we will start running programs from every other. The longer we leave it running, the more of a mixture we end up with. Since Chaitin's Constant is the (uncomputable) result of letting such a scheme run forever, it contains a perfect mixture of all instruction sets, and is thus independent of whichever one we choose.