Place all the 5s. The key one is the hidden single in R6C5. Then a few more hidden singles.
Slitherlink deductions based on shaded numbers.
R3C2 can't be a 0 clue in the Slitherlink since it must have an edge in the loop. That force it to be a 1. Several more hidden singles follow.
A 0 Slitherlink clue can't border a 3 in a corner, therefore 6 can't go in R5C6. Now C6 can be filled in with hidden singles.
Some Slitherlink deductions with the new shaded numbers.
Either way the 3 in a corner goes, the border between R5C5 and C5C6 must be used. Therefore R5C5 can't be a 0 clue, forcing it to be a 2. I also added in some connectivity Xs.
If R1C2 was either a 3 or a 0 clue, then the border between R1C2 and R1C3 couldn't be used. So it isn't. Then the border between R2C2 and R2C3 must be used, or bad things happen quickly - namely, R1C4 is forced to be a 2, which... doesn't work.
The number of ends entering any one area must be even so that they can match up for the loop. Therefore the 2 in R3C4 must use parallel borders, to avoid creating three hanging ends for the top area. Either way the borders between R2C3-4-5 must all be used.
If the border between R1C3 and R2C3 is used then the 2 in R1C3 can't be satisfied as a Slitherlink clue. Adding an X there allows the top part of the loop to be resolved.
The loop segments at the top give some Sudoku numbers, and with them and some hidden singles the Sudoku grid is resolved.
Some more Slitherlink deductions with the new shaded numbers.
Trying to make the 3 in the corner use its top edge quickly runs into problems, so it doesn't use its top edge. (Note: I really wish I didn't case-bash here but I couldn't figure out any other way to proceed. I'm open to any suggestions). Some quick deductions arise from that.
Placing some more Xs reveals which side of each 1 must be used.
Remembering that we can't close the loop too early, the rest is simple.