I think Ady's question has a negative answer.

Recall that the Mazur map, $T$, from the unit ball of $\ell_2$ onto the unit ball of $\ell_1$, is defined by
$T(\sum a_i e_i)= \sum a_i^2 b_i e_i$, where $b_i$ is the sign of $a_i$. It is a uniform homeomorphism in the norm topologies and obviously is coordinatewise continuous, which means that it is weak ($=$ weak$*$ in $\ell_2$) to weak$^*$ continuous (since on the unit ball of $\ell_p$ with $p$ finite the
weak* topology is the topology of coordinatewise convergence).
$T$ is not weak to weak continuous since the unit vector basis converges weakly to zero in $\ell_2$ but not in $\ell_1$.

The problem is to extend $T$ to all of $\ell_2$. I think that the easiest way to do this is to show that there is a retraction $R$ from $\ell_2$ onto its unit ball which is both norm to norm continuous and weak to weak continuous. Define $R$ on the complement of the unit ball by

$$
R(\sum a_i e_i) = \sum_{i=1}^n a_i e_i + t e_{n+1},
$$
where $\sum_{i=1}^n a_i^2 \le 1 < \sum_{i=1}^{n+1} a_i^2$ and $t$ is chosen to have the same sign as $a_{n+1}$ and to make the image vector have norm one. $R$ is obviously continuous (even Lipschitz) in the norm topology and is continuous in the topology of coordinatewise convergence, hence is weak to weak continuous (since in all of $\ell_2$, the weak topology is stronger than the topology of coordinatewise convergence).

To get a counterexample that maps one dual space to itself, work with the space $\ell_2 \oplus \ell_1$.