Before I answer, let me comment that duals of von Neumann algebras are highly pathological objects, and when you get to that level I suspect most operator algebraists would see you as doing set theory, not operator algebras. Just FYI.
The answer is no, by the usual sort of counterexample. Let $M = B(l^2)$ and let $(e_n)$ be the standard basis of $l^2$. For each $n$ let $a_n$ be the operator $v \mapsto \langle v, e_1\rangle e_n$ and let $f_n$ be the linear functional $a \mapsto \langle ae_1, e_n\rangle$. Then $a_nf_n(I) = f_n(a_n) = 1$, so $a_nf_n \not\to 0$ weak*. However, $f_n \to 0$ weak* and $a_n \to 0$ weakly. Both of these claims follow from the fact that the "first column" operator space (all $a \in B(H)$ whose range is contained in ${\rm span}(e_1)$) is isometrically isomorphic to $l^2$. Thus $f_n \to 0$ weak* because $f_n(a) $ reads off the $n$th entry of the first column of $a$, which goes to $0$ since the column belongs to $l^2$. And $a_n \to 0$ weakly because it lies in a subspace of $B(l^2)$ that is isometrically isomorphic to $l^2$, where it corresponds to $e_n \in l^2$, and the sequence $(e_n)$ goes to $0$ weakly in $l^2$.