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
$0$ is an element of the weak closure of $I$: fix a basic neighborhood of $0$ for the weak topology, say $O:=\bigcap_{j=1}^N\{x,|f_j(x)|<\delta\}$ where $\delta>0$ and $f_j$ are linear continuous functionals. We have to show that $O$ contains an element $\sqrt n\cdot e_n$ for some $n$. If not, representing the linear functional $f_j$ by the vector $y_j$, we would have $$\sum_{j=1}^N\|y_j\|^2=\sum_{j=1}^N\sum_{n=1}^\infty|\langle e_n,y_{j}\rangle|^2=\sum_{n=1}^\infty\sum_{j=1}^N|\langle e_n,y_{j}\rangle|^2\geqslant\sum_{n=1}^\infty\frac{\delta^2}{n}=\infty.$$
Let $\{x_n\}$ a sequence of $I$. If there are infinitely many different terms, say $\{\sqrt k\,e_k,\ k\in A\}$ where $A\subset\Bbb N$ is infinite, write the sequence $x_k:=\sqrt{n_k}e_{n_k}$ where $n_k$ is an increasing sequence of integers. The sequence $\{x_k\}$ is not bounded an so cannot be weakly convergent. If there are only finitely many different terms, we extract a subsequence constant equal to one of them, proving we can't have convergence to $0$.
If there were a decreasing countable basis of neighborhood at $0$ , say $\{V_n,n\in\Bbb N\}$ (for the weak topology), we would be able for each $n$, $x_{k_n}\in I\cap V_n$ by the first point (and the definition of the closure). And this sequence would converge weakly to $0$, a contradiction by the second item of the list.
A metric space $(S,d)$ satisfies the first axiom of countability, as if $x\in S$, the collection $\{B_d(x,n^{-1}),n\in\Bbb N^*\}$ would be a countable basis of neighborhoods.