Let's begin with a more abstract viewpoint.
In every topological space $X$, for a subset $A \subset X$, we have the characterisation of the closure of $A$ that $x$ belongs to $\overline{A}$ if and only if there is a filter $\mathscr{F}$ on $A$ whose image under the inclusion $\iota\colon A \hookrightarrow X$ converges to $x$. That is rather trivial - if $x\in \overline{A}$, the trace of its neighbourhood filter $\mathscr{V}(x)$ on $A$ is a filter whose image under $\iota$ is finer than $\mathscr{V}(x)$, and if $\iota(\mathscr{F}) \to x$, then for every neighbourhood $V$ of $x$ there is some $F\in \mathscr{F}$ with $F\subset V$, whence $V\cap A \supset F \neq \varnothing$ - but useful.
Now convergence is preserved when we go from one filter to a finer filter, and every filter is contained in an ultrafilter, so we can also characterise the closure using ultrafilters:
$x\in \overline{A}$ if and only if there is an ultrafilter $\mathscr{U}$ on $A$ with $\iota(\mathscr{U}) \to x$.
We note that the image of an ultrafilter is always an ultrafilter.
Now we come to the situation of the question. Let $E = \{ e_n : n \in \mathbb{N}\}$ and $B$ the norm-closed unit ball of $(\ell^{\infty})^{\ast}$. By Banach-Alaoglu-Bourbaki, we know that $B$ is $\omega^{\ast}$-compact, so every ultrafilter on $B$ converges. Since the inclusion $\iota \colon E \hookrightarrow (\ell^{\infty})^{\ast}$ factors through $B$, $\iota(\mathscr{U})$ is convergent for every ultrafilter $\mathscr{U}$ on $E$. With the obvious bijection $e \colon \mathbb{N} \to E$, we can identify ultrafilters on $E$ and ultrafilters on $\mathbb{N}$, thus we have a natural surjection
$$\lambda \colon \operatorname{Ult}(\mathbb{N}) \twoheadrightarrow \overline{E}^{\omega^{\ast}};\quad \lambda(\mathscr{U}) = \lim \iota(\mathscr{U}).$$
In fact, $\lambda$ is a bijection. Let $\mathscr{U}_1,\mathscr{U}_2$ two different ultrafilters on $\mathbb{N}$, and $f_k = \lim \iota(\mathscr{U}_k)$, $k\in \{1,2\}$. Since $\mathscr{U}_1 \neq \mathscr{U}_2$, there is an $F\subset \mathbb{N}$ with $F\in \mathscr{U}_1$ and $\mathbb{N}\setminus F \in \mathscr{U}_2$. Then $\langle 1_F, e_n\rangle = 1$ for all $n\in F$, whence $f_1(1_F) = 1$, and $\langle 1_F, e_n\rangle = 0$ for all $n \in \mathbb{N}\setminus F$, whence $f_2(1_F) = 0$, so $f_1 \neq f_2$, and the injectivity of $\lambda$ is proved.
Thus we have the natural bijection $\lambda \colon \operatorname{Ult}(\mathbb{N}) \to \overline{E}^{\omega^{\ast}}$, and one verifies without problems that indeed
$$\lambda(\mathscr{U})(x) = \lim_{\mathscr{U}} x$$
for all $x\in \ell^{\infty}$.
A more concrete argument proceeds as follows:
We again let $E = \{ e_n : n \in \mathbb{N}\}$, and write $C = \overline{E}^{\omega^{\ast}}$.
For all $f\in C$ and $A \subset \mathbb{N}$ we have $f(1_A) \in \{0,1\}$. For if $f \in (\ell^{\infty})^{\ast}$ with $f(1_A) \notin \{0,1\}$, the $\omega^{\ast}$-neighbourhood $V(f;1_A;\varepsilon) = \{g\in (\ell^{\infty})^{\ast} : \lvert f(1_A) - g(1_A)\rvert < \varepsilon\}$ of $f$ contains no $e_n$ if $\varepsilon \leqslant \min \{ \lvert f(1_A)\rvert, \lvert 1 - f(1_A)\rvert\}$, so then $f \notin C$.
$f(1_{\mathbb{N}}) = 1$ for all $f\in C$. That follows immediately from $e_n(1_{\mathbb{N}}) = 1$ for all $n\in \mathbb{N}$.
For $f\in C$ and $A \subset \mathbb{N}$, we have either $f(1_A) = 1$ and $f(1_{\mathbb{N}\setminus A}) = 0$ or $f(1_A) = 0$ and $f(1_{\mathbb{N}\setminus A}) = 1$. That follows from $1_{\mathbb{N}} = 1_A + 1_{\mathbb{N}\setminus A}$ and the first two points.
If $f\in C$, $A\subset \mathbb{N}$ with $f(1_A) = 1$ and $B \supset A$, then $f(1_B) = 1$. Since $f\in C$, the $\omega^{\ast}$-neighbourhood $V\bigl(f;1_A,1_B;\frac{1}{2}\bigr)$ contains some $e_n$. From $\lvert f(1_A) - e_n(1_A)\rvert < \frac{1}{2}$, it follows that $n \in A \subset B$, and then from $\lvert e_n(1_B) - f(1_B)\rvert = \lvert 1 - f(1_B)\rvert < \frac{1}{2}$ and point 1, we deduce $f(1_B) = 1$.
If $f\in C$ and $A,B \subset \mathbb{N}$ with $f(1_A) = 1 = f(1_B)$, then $f(1_{A\cap B}) = 1$. Since $f\in C$, the $\omega^{\ast}$-neighbourhood $V\bigl(f; 1_A, 1_B, 1_{A\cap B};\frac{1}{2}\bigr)$ contains some $e_n$. Evaluating at $1_A$ and $1_B$ shows that $n \in A\cap B$, and then evaluating at $1_{A\cap B}$ shows $\lvert 1 - f(1_{A\cap B})\rvert < \frac{1}{2}$, whence per point 1 we have $f(1_{A\cap B}) = 1$.
Items 1. to 5. show that for every $f \in C$ the family
$$\mathscr{U}_f = \{ A \subset \mathbb{N} : f(1_A) = 1\}$$
is an ultrafilter on $\mathbb{N}$. (Side remark: the map $f \mapsto \mathscr{U}_f$ is of course the inverse of the bijection $\lambda$ from above.)
Now one uses that $\operatorname{span} \:\{ 1_A : A \subset \mathbb{N}\}$ is dense in $\ell^{\infty}$ to deduce that $f \mapsto \mathscr{U}_f$ is injective, and that for all $x \in \ell^{\infty}$
$$f(x) = \lim_{\mathscr{U}_f} x.$$
It remains to be seen that $f \mapsto \mathscr{U}_f$ is surjective, or equivalently, that $\lim_{\mathscr{U}} \in C$ for every ultrafilter $\mathscr{U}$ on $\mathbb{N}$. Of course that follows immediately from the first part of this answer, but in the spirit of this second part, let's give a more concrete argument. Let $\varepsilon > 0$ and $x_1,\dotsc x_r \in \ell^{\infty}$ be given. We need to show that $V(f; x_1,\dotsc, x_r;\varepsilon) \cap E \neq \varnothing$, where $f = \lim_{\mathscr{U}}$.
For $1 \leqslant \rho \leqslant r$, let $c_{\rho} = f(x_{\rho})$, and $A_{\rho} = \{ n \in \mathbb{N} : \lvert x_{\rho}(n) - c_{\rho}\rvert < \varepsilon\}$. Then $A_{\rho} \in \mathscr{U}$, and since $\mathscr{U}$ is a filter, we have $A = A_1 \cap \dotsc \cap A_r \in \mathscr{U}$. In particular, $A \neq \varnothing$. But by definition,
$$V(f;x_1,\dotsc,x_r;\varepsilon) \cap E = \{ e_n : n \in A\}.$$
