It is not sequential.

See the paper

> Humphrey, A. James and Simpson, Stephen G.  Separable Banach space theory needs strong set existence axioms.  *Trans. Amer. Math. Soc.* **348** (10), 4231-4255, 1996.  [Open access full text](http://www.ams.org/journals/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pdf)

Theorem 2.5 of that paper shows that for any infinite-dimensional separable Banach space $X$, there is a countable subset $Z \subset X^*$ which is weak-* sequentially closed but weak-* dense (in particular, not weak-* closed).  I believe your definition of sequential is equivalent to "every sequentially closed set is closed" ([Wikipedia says so](http://en.wikipedia.org/wiki/Sequential_space) but does not give the proof), so this shows $X^*$ is not weak-* sequential.

Of course, if we take $X = C([0,1])$ (which is separable) then by the Riesz representation theorem, we have $X^* = M([0,1])$, so $M([0,1])$ is not sequential in the weak-* topology.