No, not necessarily. Any reflexive Banach space has a weakly compact unit ball; so, by the Eberlein-Šmulian Theorem, any reflexive Banach space has a weak* sequentially compact unit ball.
Also, and more generally, it follows from Rosenthal's $\ell_1$ Theorem (a Banach space $X$ does not contain $\ell_1$ isomorphically if and only if every every bounded sequence in $X$ has a weakly Cauchy subsequence) that if $X^*$ does not contain $\ell_1$ isomorphically, then the unit ball of $X^*$ is weak* sequentially compact.
No, it cannot be sequential unless $X$ is finite-dimensional. Otherwise, for each $k$ we may pick a subspace $X_k$ of $X$ with $\dim X_k = k$. Moreover, we choose a finite $\tfrac{1}{k}$-net $x_{k,j}$ of the sphere $\{x\in X_k\colon \|x\|=k\}$ in $X_k$ (possible by compactness). Let $S$ be the union of all the nets picked above. We claim that 0 is in the weak closure of $S$.
Indeed, let $U$ be a weakly open neighbourhood of 0. Let $f_1, \ldots, f_n\in X^*$ be norm-one functionals and let $\varepsilon > 0$ be such that
$$\{x\in X\colon \max_i |\langle f_i, x\rangle| < \varepsilon \}\subseteq U.$$
Take $k$ with $1/k <\varepsilon$. When $n<k$, there must be $y_k\in X_k$ such that $\langle f_i, y_k\rangle = 0$ for all $i$. Without loss of generality $\|y_k\|=k$. Pick $j$ so that $\|x_{k,j} - y_k\|\leqslant\tfrac{1}{k}$. Consequently,
$$|\langle f_i, x_{k,j}\rangle| = |\langle f_i, x_{k,j} - y_k\rangle| \leqslant \|x_{k,j} - y_k\|\leqslant\tfrac{1}{k}<\varepsilon, $$
that is $x_{k,j}\in U$.
This establishes the claim and thus $S$ is not weakly closed.
On the other hand, every weakly convergent sequence in $S$ is bounded, and thus lives only on finitely many points of $S$. Hence, the weak limit belongs to $S$. This yields that $S$ is weakly sequentially closed.
There is a strengthening of this result by Gabriyelyan, Kąkol and Plebanek (see Theorem 1.5 here):
Theorem. Let $E$ be a Banach space. Then the weak topology of $E$ has the Ascoli property if and only if $E$ is finite-dimensional.
Best Answer
Let $JT$ denote the James Tree space, $Q: JT \longrightarrow JT\,^{\prime\prime}$ the canonical embedding, $D_\ast$ a countable norm-dense subset of $JT$, let $D= Q(D_\ast)$ and let $E = JT\,^\prime$. Note that $E$ is nonseparable in the norm topology and that the $w^\ast$-sequential closure of $D$ in $E^{\prime}$ is $E^{\prime}$ since $Q(JT)$ is $w^\ast$-sequentially dense in $JT\,^{\prime\prime}$ (for the last claim, see in particular Corollary 2 of Lindenstrauss and Stegall Examples of separable spaces which do not contain $\ell_1$ and whose duals are non-separable, Studia Math. 54 (1975), p.81--105).
I should mention that until the appearance of the Lindenstrauss-Stegall result cited above, it seems to have been open since the time of Banach whether there could exist a separable Banach space that has nonseparable dual and is $w^\ast$-sequentially dense in its bidual.