Maybe there's some other technical obstacle I'm missing here

There's not. The Hartog's number construction gives us the set H(N) of all isomorphism classes of well-orders on subsets of any fixed countably infinite set, and we can prove that H(N) is uncountable and every proper initial segment of H(N) is countable, using power set and separation (but only bounded separation) but not replacement. I verified this just now by looking at Wikipedia's article on Hartog's number and checking through the proof myself.

The next step (step 4 in Wikipedia, ETA: which can be saved for the end, although WP did not do so) is to replace the elements of H(N) with von Neumann ordinals, but this is really beside the point. You already have a representation of the least uncountable ordinal, and this step is just making it canonical in a certain way.