It is not sequential. Here is an explicit counterexample.
Notation: $\|\mu\|$ is the total variation norm of $\mu$, and $\bar{B}_r \subset M([0,1])$ is the closed norm ball of radius $r$ centered at $0$. $\delta_x$ is the Dirac measure which places a unit point mass at $x \in [0,1]$.
For each positive integer $n > 0$, let $P_n \subset [0,1]$ be any finite set of size at least $(2n)^{2n}$, and let $$E_n = \{ n(\delta_{x} - \delta_y) : x,y \in P_n, \, x \ne y\} \subset M([0,1]).$$
Notice that $E_n$ is finite, and that $\|\mu\| = 2n$ for every $\mu \in E_n$.
Set $E = \bigcup_{n=1}^{\infty} E_n$. I claim that 0 is a weak-* limit point of $E$. Let $f_1, \dots, f_m \in C([0,1])$ and let $\epsilon > 0$. I will produce $\mu \in E$ with $\left|\int f_k\,d\mu\right| < \epsilon$ for every $k$. Without loss of generality, assume $\|f_k\|_\infty \le 1$ for all $k$ (otherwise divide $\epsilon$ by $\max_k \|f_k\|_\infty$). Define $F : [0,1] \to [-1,1]^m$ by $F = (f_1, \dots, f_m)$. Choose $n > \max(m, 1/\epsilon)$.
We can cover $[-1,1]^m$ with $(2n^2)^{m}$ cubes of side length $1/n^2$. Since $$|P_n| \ge (2n)^{2n} > (2n)^{2m} = (4n^2)^m > (2n^2)^m$$ by the pigeonhole principle there must exist two distinct $x,y \in P_n$ with $F(x), F(y)$ in the same cube. This means that $|f_k(x) - f_k(y)| \le 1/n^2$ for every $k$. So if we take $\mu = n(\delta_x - \delta_y) \in E_n$, we have $\left|\int f_k\,d\mu\right| \le 1/n < \epsilon$ for every $k$. This proves the claim that 0 is a weak-* limit point of $E$.
Now for any $n$, we know that $\bar{B}_n$ is weak-* closed and disjoint from the finite set $E_n$. The weak-* topology is completely regular, so there exists a weak-* continuous function $G_n : M([0,1]) \to [0,1]$ with $G_n = 0$ on $\bar{B}_n$ and $G_n = 1$ on $E_n$. Set $G = \sum_{n=1}^\infty G_n$. Note that on any ball $\bar{B}_n$, only finitely many terms of the sum are nonzero, so the sum makes sense, and the restriction of $G$ to any ball is weak-* continuous. If $\mu_k$ is a weak-* convergent sequence with limit $\mu$, then by the uniform boundedness principle all the $\mu_k$ and $\mu$ lie in some ball $B$. The restriction of $G$ to $B$ is weak-* continuous, so $G(\mu_k) \to G(\mu)$. Thus we have shown $G$ is weak-* sequentially continuous.
On the other hand, $G$ is not weak-* continuous, since $G(0) = 0$ but $G \ge 1$ on $E$.
(The same uniform boundedness argument shows that $E$ is weak-* sequentially closed, but not weak-* closed since it does not contain 0.)
More generally, we can replace $M([0,1])$ by the dual of any separable Banach space. 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, $Z$ is not weak-* closed, so the weak-* topology on $X^*$ is not sequential in the usual sense of the word.
To show it is not sequential in your sense: As an intermediate step of their construction, they get a sequence which has $0$ as a weak-* limit point but intersects every ball in only finitely many points; we could proceed as we did above to construct a function $G : X^* \to \mathbb{R}$ which is weak-* sequentially continuous but not weak-* continuous. (Their set $Z$ is formed by translating such a sequence by a countable weak-* dense set.)
