I am going to assume that by an additive character you mean
an irreducible representation $\chi_\alpha : \mathbb{F}^n_q \longrightarrow \mathbb{C}$, i.e. a group homomorphism from the additive group $(\mathbb{F}^n_q ,+)$ to the multiplicative group $(\mathbb{C},*)$
which we can prove must all take the form \begin{equation}\chi_\alpha : \beta \mapsto \exp\left( {\frac{2\pi i \left\langle \alpha ,\beta \right\rangle }{p }} \right)\end{equation} where $ \left\langle \alpha ,\beta \right\rangle = \sum_i \alpha_i \beta_i $, see chapter 4 of Tao for a proof of some of these statements and see ch.2 of Serre or ch.2 of Fulton & Harris for a general (non-abelian) overview of the representation theory perspective on characters. The point is the following
If we let \begin{equation} f(x) = \begin{cases} q \psi_x(x) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \\ \end{cases} \end{equation} then the sum you are considering is equal to the Fourier transform of $f$ i.e. \begin{equation} \hat{f}(\alpha) = \frac{1}{q} \sum_{c \in \mathbb{F} _q } f(c) \chi_\alpha(c) = \sum_{c \in \mathbb{F} _q^* } \psi_c (c) \chi_\alpha(c) \end{equation} see definition 4.6 in Tao.
We apply the Hausdorff-Young inequality theorem 4.8 in Tao to get that \begin{equation} \left(\sum_{\alpha \in \mathbb{F} _q }\left| \hat f(\alpha)\right|^{p'} \right)^{\frac{1}{p'}} \leq \left(\sum_{\alpha \in \mathbb{F} _q } |f(\alpha)|^p\right)^{\frac{1}{p}} = q\left( \sum_{c \in \mathbb{F} _q^* } |\psi_c (c) |^p\right)^{\frac{1}{p}} \end{equation} where the LHS is the $l^{q}$-norm, the RHS is the $l^p$-norm, and $p$ satisfies the following $p^{-1} +q^{-1} = 1 \land 1 \leq p\leq 2$. Plugging in $p = 2$ we get that
\begin{equation} \sum_{\alpha \in \mathbb{F} _q }\left| \hat f(\alpha)\right|^{2} \leq q\sum_{c \in \mathbb{F} _q^* } |\psi_c (c) |^2 \end{equation} which is equivalent to saying that \begin{equation} \mathbb{Var}[\hat f] = \frac{1}{q}\sum_{\alpha \in \mathbb{F} _q }\left| \hat f(\alpha)\right|^{2} \leq \sum_{c \in \mathbb{F} _q^* } |\psi_c (c) |^2\leq q-1. \end{equation}
Finally, if you can prove that at least $n$ many $\alpha$ give a value $ \sqrt b\leq | \hat f(a)| \leq \sqrt {b+ \epsilon }$ then you get that \begin{equation} nb +\sup_{\alpha \in \mathbb{F} _q }\left| \hat f(a)\right|^2 = \sum_{\alpha \in S}\left| \hat f(\alpha)\right|^{2} + \sup_{\alpha \in \mathbb{F} _q }\left| \hat f(a)\right|^2 \leq \sum_{\alpha \in \mathbb{F} _q }\left| \hat f(\alpha)\right|^{2} \leq q-1 \end{equation}
which gives you that the maximum value is at most
\begin{equation} \sup_{\alpha \in \mathbb{F} _q }\left|\sum_{c \in \mathbb{F} _q^* } \psi_c (c) \chi_\alpha(c) \right| = \sup_{\alpha \in \mathbb{F} _q }\left| \hat f(a)\right| \leq \sqrt{q-1-nb} \end{equation}
Essentially we reduced the problem of finding an upper bound to that of finding a lower bound.