Is there any known example of fibered knot which is topologically slice but not (expected to be) smoothly slice?

$\begingroup$ No knot with 12 crossings or fewer has this property, according to knotinfo. Anyway, is there any particular reason you're looking for such an example? $\endgroup$– Marco GollaAug 24 '15 at 12:25

$\begingroup$ Freedman's work (mentioned in Jim Conant's answer) gives that Alexander polynomial 1 knots are topologically slice. But, as mentioned by Dan, Alexander polynomial 1 knots are not fibered. I was curious that there are any known construction of fibered topologically slice knot which 'uses' Freedman's disk embedding theorem. $\endgroup$– user156937Aug 24 '15 at 14:03
Such a knot would yield a counterexample to one of two important conjectures in the area. A preliminary definition: a slice knot is homotopically ribbon if the inclusion of the knot into the slice disk complement induces a surjection on the fundamental group. It's easy to see that a ribbon knot is homotopically ribbon; note that homotopically ribbon doesn't require any smoothness (but you still want local flatness to avoid trivialities). This criterion is viewed as the right substitute for ribbon in the topological setting.
The remarkable theorem of CassonGordon (A loop theorem for duality spaces and fibred ribbon knots; Invent. Math. 74, 119137 (1983)) says that a fibered knot is homotopically ribbon if and only if its monodromy (filled in over a disk) extends over a handlebody. The if part constructs a smooth slice smooth homotopy ball with boundary the 3sphere (if your knot lay in the 3sphere.) So in particular, a fibered knot that is topologically homotopy ribbon is smoothly homotopically ribbon!!! So an example of the sort that you ask for would either be slice but not homotopically ribbon (contradicting the conjecture that this doesn't happen) or would give a smooth slice counterexample to the 4dimensional Poincaré conjecture.
So to summarize, no such example is known.
PS If there were an MO question asking for "your favorite paper that people don't think about enough", the CassonGordon paper mentioned above might top my list.
A common source of topologically slice knots are those with Alexander polynomial $1$. However these are not fibered. This follows from a classical result, that $2\mathrm{genus}(K) = \mathrm{deg}(\Delta_K(t))$ for fibered knots, which I learned in a paper of Stefan Friedl and Taehee Kim ("The Thurston Norm, Fibered Manifolds and Twisted Alexander Polynomials"). Of course there are other sources of topologically slice knots, so this doesn't settle the question.

1$\begingroup$ Your reference is an overkill. A fibering gives you an infinite cyclic cover with homotopy type of the fiber. The characteristic polynomial of $H_1($generator of the Deck transformations$;\mathbb Q) \in Aut(H_1(F;\mathbb Q) \cong \mathbb Q^{2g})$ is the Alexander polynomial. Hence the equality. $\endgroup$ Aug 24 '15 at 13:39