If $\alpha_1,\ldots,\alpha_{2g}$ are the eigenvalues on $H^1(A,\mathbb Q_\ell)$, then $\bigwedge^* H^1 = H^*$ computes all eigenvalues. Moreover, any $\frac{q}{\alpha_i}$ is another one of the $\alpha_i$ (see for example [Suh]). Thus, the sum $$\#A(\mathbb F_{q^r}) = \sum_{k=0}^{2g} (-1)^k\operatorname{tr}(\operatorname{Frob}_q^r|H^k(A,\mathbb Q_\ell))$$ is entirely determined by suitable algebraic expressions in $\alpha_1,\ldots,\alpha_g$ (w.l.o.g.). Thus, if we know $g+1$ of its values, then we know all of them. But we get the value for $r = 0$ for free, for this is the Euler characteristic of $A$.
[Suh] Suh, Junecue, Symmetry and parity in Frobenius action on cohomology, Compos. Math. 148, No. 1, 295-303 (2012). ZBL1258.14023.).