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Sturmfels and Zworski have an unpublished conjecture, the "Chez Panisse conjecture", which I believe is meant to say that that adding one more $N_r$ to the list will determine $f$. I say "I believe" because the published restatements I can find (Hillar's NSA proposal, Hillar and Levine, Kedlaya) state the conjecture as saying that, if $f$ is reciprocal (meaning $f(x)=x^{2g} f(1/x)$) then $f$ can be recovered from its first $g+1$ cyclic resultants (meaning $\prod_{\zeta^r=1} f(\zeta)$). But $f$ for an abelian variety obeys $f(x)=x^{2g} f(q/x)$, not $f(x)=x^{2g} f(1/x)$ and, while we could make a change of variable to work with a reciprocal $f$, I believe we would then want $\prod_{\zeta^r=1} f(q^{-1/2} \zeta)$. Since I can't find a place where Sturmfels and Zworski themselves wrote this down, I can't say whether they missed this issue, whether everyone reporting on them did, or whether they for some reason wanted to state a conjecture which wasn't quite the right one for applications to number theory.

Sturmfels and Zworski have an unpublished conjecture, the "Chez Panisse conjecture", which I believe is meant to say that that adding one more $N_r$ to the list will determine $f$. I say "I believe" because the published restatements I can find (Hillar's NSA proposal, Hillar and Levine, Kedlaya) state the conjecture as saying that, if $f$ is reciprocal (meaning $f(x)=x^{2g} f(1/x)$) then $f$ can be recovered from its first $g+1$ cyclic resultants (meaning $\prod_{\zeta^r=1} f(\zeta)$). But $f$ for an abelian variety obeys $f(x)=x^{2g} q^{-g} f(q/x)$, not $f(x)=x^{2g} f(1/x)$ and, while we could make a change of variable to work with a reciprocal $f$, I believe we would then want $\prod_{\zeta^r=1} f(q^{-1/2} \zeta)$. Since I can't find a place where Sturmfels and Zworski themselves wrote this down, I can't say whether they missed this issue, whether everyone reporting on them did, or whether they for some reason wanted to state a conjecture which wasn't quite the right one for applications to number theory.

So $f$ is given by $g$ parameters $a_1$, $a_2$, ...,, $a_g$. The theorem of Honda and Tate comes close to saying that any $a_i$ obeying (1) and (2) give the characteristic polynomial of a $g$-dimensional abelian variety. Actually, what it actually says is that any such $f$ divides the characteristic polynomial of some abelian variety, possibly of larger dimension, but in many cases (for example, whenever $a_g \not \equiv 0 \bmod q$) the dimension is $g$, so lets just look at (1) and (2).

Searching for a counter-example If I disregard (1) and (2), it is easy to give examples of nonuniqueness with $g=3$. Let $\omega$ be a primitive cube root of $1$. Find a real polynomial $h$ of the form $$h(x) = x^6+a_1 x^5+a_2 x^4+a_3 x^3 + q a_2 x^2 + q^2 a_1 x + q^3$$ such that $$\mathrm{arg}(h(\omega))+\pi/2 = \mathrm{arg}(\omega (\omega^2-1)(\omega^2-q)).$$ Choose some real scalar $t$ and set $f_{\pm}(x) = h(x) \pm t x (x^2-1)(x^2-q)$. Then $$f_+(1)=f_-(1)=h(1)$$ $$f_+(-1)=f_-(-1)=h(-1)$$ $$f_+(\omega) f_+(\bar{\omega}) = |h(\omega)| + t^2 |\omega (\omega^2-1)(\omega^2-q)|^2 = f_-(\omega) f_-(\bar{\omega}).$$ So $f_+(1)=f_-(1)$, $f_+(-1)=f_-(-1)$ and $f_+(\omega) f_+(\bar{\omega})=f_-(\omega) f_-(\bar{\omega})$ but $f_+ \neq f_-$.

So $f$ is given by $g$ parameters $a_1$, $a_2$, ...,, $a_g$. The theorem of Honda and Tate comes close to saying that any $a_i$ obeying (1) and (2) give the characteristic polynomial of a $g$-dimensional abelian variety. Actually, what it actually says is that any such $f$ divides the characteristic polynomial of some abelian variety, possibly of larger dimension, but in many cases (for example, whenever $a_g \not \equiv 0 \bmod p$) the dimension is $g$, so lets just look at (1) and (2).

Searching for a counter-example If I disregard (1) and (2), it is easy to give examples of nonuniqueness with $g=3$. Let $\omega$ be a primitive cube root of $1$. Find a real polynomial $h$ of the form $$h(x) = x^6+a_1 x^5+a_2 x^4+a_3 x^3 + q a_2 x^2 + q^2 a_1 x + q^3$$ such that $$\mathrm{arg}(h(\omega))+\pi/2 = \mathrm{arg}(\omega (\omega^2-1)(\omega^2-q)).$$ Choose some real scalar $t$ and set $f_{\pm}(x) = h(x) \pm t x (x^2-1)(x^2-q)$. Then $$f_+(1)=f_-(1)=h(1)$$ $$f_+(-1)=f_-(-1)=h(-1)$$ $$f_+(\omega) f_+(\bar{\omega}) = |h(\omega)|^2 + t^2 |\omega (\omega^2-1)(\omega^2-q)|^2 = f_-(\omega) f_-(\bar{\omega}).$$ So $f_+(1)=f_-(1)$, $f_+(-1)=f_-(-1)$ and $f_+(\omega) f_+(\bar{\omega})=f_-(\omega) f_-(\bar{\omega})$ but $f_+ \neq f_-$.

As the OP notes, for $g=1$, the quantity $f(1) = q+1-a_1$ determines $a_1$. Also, when $g=2$, the quantities $N'_1 = f(1) = q^2+1 + (q+1) a_1 + a_2$ and $N'_2 =N_2/N_1 = f(-1) = q^2+1 - (q+1) a_1 + a_2$ determine $a_1$ and $a_2$. (Since $A$ is an abelian variety, it has an identity in $A(\mathbb{F}_q)$, so $N_1 \geq 1$ and we may divide by it.) So the first interesting case is $g=3$.