As noted in two comments above, $\mathbb Q^{\mathbb N}$ is not countable and it is not a subset of $\ell^\infty$ since it contains unbounded items.
The set $T=\mathbb Q^{<\omega}$ - the set of sequences of rationals which are zero for all but finitely many values - is both countable and a subset of $\ell^\infty$. $\operatorname{cl}(T)$ is the set of elements of $\ell^\infty$ which converge to $0$ - that is, given $a=\{a_i\}_{i\in\mathbb N}\in\ell^\infty$, $a\in \operatorname{cl}(T)$ if and only if $\lim_{i\to\infty} a_i = 0$
Proof. "If." Assume $\lim_{i\to\infty} a_i =0$ and let $\epsilon>0$. Then choose $N$ such that $|a_i|<\epsilon$ for every $i>N$. Define $b_i=0$ for $i>N$, and, for $i\leq N$, find rational $b_i$ such that $|b_i-a_i|<\epsilon$. Then $b=\{b_i\}\in T$, and $|b_i-a_i|<\epsilon$ for all $i$, so $|b-a|<\epsilon$. Therefore, $a=\{a_i\}$ is in the closure of $T$.
"Only if." On the other hand, assume $a=\{a_i\}\in \operatorname{cl}(T)$. Let $\epsilon>0$. Then, by the definition of closure, there exists $b=\{b_i\}\in T$ such that $|b-a|<\epsilon$. Let $N$ be a number for which $b_i=0$ for all $i>N$ (which exists since $b\in T$.) Then $|a_i-b_i|=|a_i|<\epsilon$ for all $i>N$. But we've just show that for all $\epsilon>0$ there is an $N$ such that for all $i>N$ $|a_i|<\epsilon$. So $\lim_{i\to\infty}a_i = 0$.