You can find a bounded linear functional in $(\ell^\infty)^* $ that does not correspond to an element of $\ell^1$ in the following way:

Consider the following linear subspace of $\ell^\infty$:
$$W=\{a=(a_1,a_2,\dots)\in \ell^\infty : \lim_{n\to\infty }a_n\textrm{ exists}\}$$
and define the linear functional $\lambda:W\to \mathbb R$ by
$$\lambda (a)=\lim_{n\to\infty } a_n$$
it is easy to see that this is bounded and $\|\lambda\|\leq1$. Now, using the Hahn- Banach theorem extend this functional to a bounded linear functional on $\ell^\infty$, such that
$$\lambda(a)=\lim_{n\to\infty} a_n$$
for $a\in W$. This extended functional does not correspond to any element $b\in \ell^1$. If it did, then we would be able to write
$$\lambda(a)= \sum_{n=1}^\infty a_nb_n$$
for all $a\in \ell^\infty$. However, if you test $e_n=(0,\dots,0,1,0,\dots)\in \ell^\infty$ you get $\lambda(e_n)=0$ using the first definition, and $\lambda(e_n)=b_n$, using the second. This implies that $b_n=0$ for all $n$, therefore $\lambda=0$. That is a contradiction because $\lambda$ is a non-trivial functional. For example, if $a=(1,1,\dots)$ then $\lambda(a)=\lim_{n\to\infty}a_n=1$.