We know that $$G_n = \text{Gal}({\mathbb{F}}_{q^n} / \mathbb{F}_q) \cong \mathbb Z / n \mathbb Z,$$ where the Frobenius $\sigma_n$ of $\mathbb F_{q^n}$ is mapped to $1 \in \mathbb Z / n \mathbb Z$.
$G := \text{Gal}(\bar{\mathbb{F}}_q / \mathbb{F}_q)$ is the inverse limit of the $G_n$, which is $\hat {\mathbb Z} \cong \lim_n \mathbb Z / n\mathbb Z \cong \prod_p \mathbb Z_p$. $G$ is topologically generated by $\sigma$, the Frobenius on $\bar{\mathbb F}_q$, i.e. $\overline{\langle\sigma\rangle} = G$.
Now for the question: what does $G\backslash\langle\sigma\rangle$ actually contain?
I'll assume that $q = p$ is a prime for simplicity of notation. Let $l$ be an another prime and note that we have $$ G_{l}:=\lim_{n}\mathrm{Gal}({\mathbb{F}}_{p^{l^{n}}}/\mathbb{F}_{p}) = \lim_{n} \mathbb{Z}/l^{n}\mathbb{Z} = \mathbb{Z}_{l} $$ which corresponds to $l$-part of $\hat{\mathbb{Z}} = \prod_{p}\mathbb{Z}_{p}$. Choose any element $\alpha = a_{0} + a_{1}l + a_{2} l^{2} + \cdots \in \mathbb{Z}_{l} \backslash \mathbb{Z}$, then the elements corresponds to $\mu \in G_{l}$ that act as $\mathrm{Frob}_{l}^{\alpha_{k}}$ on $\mathbb{F}_{p^{l^{k}}}$ with $\alpha_{k} = a_{0} + a_{1}l + \cdots + a_{k}l^{k}$ for all $k\geq 0$. We can lift $\mu$ to $\tilde{\mu} \in G$ under the surjective map $G \twoheadrightarrow G_{l}$, and that element is not in $\langle \sigma\rangle$ because of the choice of $\alpha$ ($\alpha \not \in \mathbb{Z}$).