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
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.