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
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 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.