Every value $c \in \big[0,\tfrac{1}{2}\big] \cap \mathbf Q$ can occur as the smallest slope of an abelian variety over $\mathbf F_q$; see the corollary below.
What Honda actually proves [Hon68] (see [Mil94, Prop. 2.6] for a motivic reinterpretation) is:
Theorem (Honda). Let $q$ be a power of a prime $p$. Then the map
\begin{align*}
\frac{\{\text{simple abelian varieties } A \text{ over } \mathbf F_q\}}{\{\mathbf F_q\text{-isogeny}\}} &\to \frac{\{q\text{-Weil numbers}\}}{\text{conjugacy}}\\
A &\mapsto \operatorname{Frob}_A
\end{align*}
is a bijection.
In other words, every $q$-Weil number (of weight $1$) is realised inside some abelian variety; in particular inside some smooth projective curve (cut by smooth hyperplanes).
So the only question is: what are the $q$-Weil numbers? This is a purely number theoretic question. If you only want to know the valuation¹ of $\alpha$ and all its conjugates, it is not so hard to come up with examples.
Lemma. Let $a, b \in \mathbf Z$ be coprime with $0\leq a \leq \tfrac{b}{2}$, and let $\alpha$ be a root of
$$f(x) = x^{2b} + q^ax^b + q^b.$$
Then $\alpha$ is a $q$-Weil number with slopes
$$\big\{\underbrace{\tfrac{a}{b},\ldots,\tfrac{a}{b}}_b,\underbrace{\tfrac{b-a}{b},\ldots,\tfrac{b-a}{b}}_b\big\}.$$
Proof. If $g(x) = x^2 + q^ax + q^b$, then $\beta = \alpha^b$ is a root of $g$. Note that $g$ is irreducible over $\mathbf Q$ (even over $\mathbf R$) since
$$\Delta = q^{2a} - 4q^b < 0.$$
Hence $\beta\bar\beta = q^b$ (the constant term of $g$), so $\alpha\bar\alpha = q$. Clearly the Newton polygon of $f$ has slopes $\tfrac{a}{b}$ and $\tfrac{b-a}{a}$, both with multiplicity $b$. $\square$
In fact it's not hard to see that $f$ is irreducible when $q = p$, using Newton polygons and irreducibility of $g$, but we don't need this.
Corollary. For every $c \in \big[0,\tfrac{1}{2}\big] \cap \mathbf Q$, there exists a simple abelian variety $A$ over $\mathbf F_q$ such that the smallest slope of $H^1_{\operatorname{\acute et}}(A_{\bar{\mathbf F}_q},\mathbf Q_\ell)$ is $c$.
Proof. Write $c = \frac{a}{b}$ with $a$ and $b$ coprime, take $\alpha$ as in the lemma, and apply Honda's theorem to get a simple abelian variety over $\mathbf F_q$ with slopes the conjugates of $\alpha$. $\square$
Remark. The question which precise set of slopes can occur on smooth projective curves is a very difficult one, and this is an active area of study. For example, it is not known if for every $(g,p)$ there exists a curve $C$ of genus $g$ in characteristic $p$ such that all slopes are $\tfrac{1}{2}$ (i.e. $C$ is supersingular). If you only care about one eigenvalue and its conjugates, you can reduce to the answer above.
¹ We choose a prime above $p$ (equivalently, pick an embedding $\bar{\mathbf Q} \hookrightarrow \bar{\mathbf Q}_p$) and normalise the valuation so that $v(q) = 1$. Then studying the valuations of $\alpha$ at primes above $p$ has become studying the valuation of the conjugates of $\alpha$, which are called the slopes of $\alpha$.
References.
[Hon68] T. Honda, Isogeny classes of abelian varieties over finite fields. J. Math. Soc. Japan 20, p. 83-95 (1968). ZBL0203.53302.
[Mil94] J. S. Milne, Motives over finite fields. Motives (Seattle, WA). Proc. Symp. Pure Math. 55.1, p. 401-459 (1994). ZBL0811.14018.
