Why doesn't Cantor's diagonal argument also apply to natural numbers?

If you represent a natural number as an infinite string, the string will become identically $0$ after a certain point. If you think it through, the "diagonal argument" in this case doesn't produce a natural number; it will produce a string with infinitely many $1$s.

On the other hand, you can consider possibly infinite binary strings --- i.e. strings in which there can be infinitely many $1$; this is one way to think of the set of $2$-adic numbers, which is indeed an uncountable extension of the set of natural numbers (as one sees using the precise diagonal argument that you suggest).

The reason is simply that natural numbers have a finite representation. (In set theory, Each one is a finite number of successions from 1, where the successor of n is n+1.) Your representation scheme essentially respects this, since for any natural number (which will be on the list), after some finite number of digits it becomes all zeros. Your diagonal element will either be one of these, and so on the list, or a sequence of ones and zeros which never 'zeros out'. This string is NOT a natural number; it corresponds to nothing in your representation scheme. The reason the diagonal argument works for real numbers is that they do not have a finite representation. In set theory they are represented as the limit of an infinite series.