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 $ | \hat f(a)| \geq \sqrt b$ then you get that
\begin{equation}
nb +\sup_{\alpha \in \mathbb{F} _q }\left| \hat f(a)\right|^2 \leq \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(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(q-1)-nb}
\end{equation}

*Essentially we reduced the problem of finding an upper bound to that of finding a lower bound.*