I think it is not sequential in your sense.

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).

The usual definition of "sequential space" is "every sequentially closed set is closed", so this shows $X^*$ is not sequential.  However, I do not immediately see how to show it fails to be sequential in the sense of your definition; maybe this part is already familiar to you.

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.