The CPython compiler is capable of making frozenset constants without being explicitly asked to.  Exactly how it does so is, of course, 'hidden' from python code.  With current main:
.
>>> dis('{1,2,3}')
  1           0 BUILD_SET                0
              2 LOAD_CONST               0 (frozenset({1, 2, 3}))
              4 SET_UPDATE               1
              6 RETURN_VALUE

Suppose one wants actually wants a frozenset, not a mutable set.  'frozenset({1,2,3})' is compiled as the above followed by a frozenset call -- making an unneeded double conversion to get what already exists. 
To avoid the intermediate set, one can use a constant tuple instead.

>>> dis('frozenset((1,2,3))')
  1           0 LOAD_NAME                0 (frozenset)
              2 LOAD_CONST               0 ((1, 2, 3))
              4 CALL_FUNCTION            1
              6 RETURN_VALUE

Even nicer would be

  1           0 (frozenset({1, 2, 3}))
              2 RETURN_VALUE

'set((1,2,3))' is compiled the same as 'frozenset((1,2,3)), but frozenset does not having the option is using a more efficient display form.  I cannot think of any reason to not call frozenset during compile time when the iterable is a constant tuple.

Serhiy, I not sure how this relates to your issue 33318 and the discussion therein about stages, but it does relate to your interest in compile time constants.