Is the last step of an iterated cryptographic hash still as resistant to preimage attacks as the original hash?
If the hash function is any good, then it should behave as a "random function" (i.e. a function chosen randomly and uniformly among all possible functions). For a random function with output size $n$ bits, it is expected that nested application will follow a "rho" pattern: the sequence of successive values ultimately enters a cycle with an expected size of $2^{n/2}$; and the length of the "queue" (the values before entering the cycle) should also be close to $2^{n/2}$.

So, theoretically, recursive application weakens the function with regards to preimage resistance, but the weakening is low-bounded by $2^{n/2}$, a boundary which requires an awfully big number of invocations to be reached.

Moreover, when we select a hash function, we want to have decent resistance to collisions, and collisions can be found with effort $2^{n/2}$. So we already choose hash functions such that $2^{n/2}$ is unreachable (e.g. we use SHA-256, because $2^{128}$ is far beyond what can be realistically achieved with Earth-bound technology; see this answer for a discussion on the subject). In short words, the weakening through recursive application has no practical consequence (while the time taken to compute all the hashes improves things greatly with regards to dictionary attacks).