Both the answers of Eric Wofsey and Jochen Wengenroth are very elegant. Let me give a quite different, a more *constructive* one. (Although AC is used repeatedly.)

**Definitions.** A *generalised limit* is a bounded linear functional on $(\ell^\infty(\mathbb N))^*$, which assigns to every
convergent sequences their limit. One such limit is called *positive* if it assigns non-negative values to sequences with non-negative terms. A *Banach limit* is a positive generalised limit $\varphi$ with the property $\varphi(S\boldsymbol{a})=\varphi(\boldsymbol{a})$, where $\boldsymbol{a}=\{a_n\}$ and $S$ is the shift operator, i.e. $S\{a_n\}=\{a_{n+1}\}$.

**Facts.** Positive generalised limits are exactly the generalised limits of unit norm. Limits with respect to ultrafilters are positive generalised limits. Limits with respect to ultrafilters are exactly the generalised limits which assign to sequences subsequential limits. Also, if $A_1,\ldots, A_k$ is a partition of $\mathbb N$ and $\mathscr F$ an ultrafilter, then exactly one of the $A_i$'s belongs to $\mathscr F$.

For $A\subset\mathbb N$, define as $\chi_A=\{a_n\}$, with $a_n=1$ if $n\in A$ and $a_n=0$ if $n\in A^c$. If $\mathscr F$ is an ultrafilter, then $l_{\mathscr F}(\chi_A)=1$ iff $A\in\mathscr F$. Otherwise $l_{\mathscr F}(\chi_A)=0$. Moreover, if $\mathscr F_i$, $i=1,\ldots,k$, are distinct ultrafilters, there exist disjoint $A_1,\ldots,A_k\subset\mathscr N$, such that $A_i\in\mathscr F_i$, $i=1,\ldots,k$.

The fact that $X=\mathrm{span}\{l_{\mathscr F}:\mathscr F\in\beta\mathbb N\}$ is not dense in $(\ell(\mathbb N))^*$ shall be a consequence of the lemma:

**Lemma.** *If $\varphi$ is a Banach limit, $\mathscr F_1,\ldots,\mathscr F_k$ are distinct ultrafilters on $\mathbb N$ and $\lambda_1,\ldots,\lambda_k\in\mathbb C$, then* $$\Big\|\varphi-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}\Big\|_*=1+\sum_{j=1}^k\lvert\lambda_j\rvert.$$

*Proof.* Let $S_{m,r}=\{mj+r: j\in\mathbb N\}$. It's not hard to see that $\varphi(\chi_{S_{m,r}})=1/m$. Meanwhile, each $\mathscr F_i$ contains exactly one of the $S_{m,r}$'s, $r=0,\ldots,m-1$, and thus, for every $\varepsilon>0$, there exist $A_i\in\mathscr F_i$, $i=1,\ldots,k$, with $\varphi(A_i)<\varepsilon$. The $A_i$'s can be chosen to be disjoint, as the $\mathscr F_i$'s are distinct. Now let $B=(\bigcup_{i=1}^k A_i)^c$. Then
$$
l_{\mathscr F_i}(\chi_{A_i})=1, \quad i=1,\ldots,k\quad\text{while}\quad 1-k\varepsilon\le\varphi(\chi_B)\le 1.
$$
Define now the sequence
$$
a_n=\left\{\begin{array}{ccc} 1 & \text{if} & n\in B, \\
\frac{\overline{\lambda}_i}{\lvert\lambda_i\rvert} & \text{if} & n\in A_i.\end{array}\right.
$$
It's not hard to check that $\|\{a_n\}\|_\infty=1$, $\lambda_i l_{\mathscr F_i}(\{a_n\})=\lvert\lambda_i\rvert$, while $\lvert\varphi(\{a_n\})-1\rvert\le 2k\varepsilon$. Therefore
$$
\Big\|\varphi(\{a_n\})-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}(\{a_n\})\Big\|_*\ge\Big\lvert\varphi(\{a_n\})-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}(\{a_n\})\Big\lvert\ge 1+\sum_{j=1}^k\lvert\lambda_j\rvert-2k\varepsilon.
$$
And the Lemma follows.