The notes I was reading mentions that if $X$ is an abelian variety of dimension $g$, over a field $k$, then we have the following cohomology: $$H^i(X,\mathcal{O}_X)=k^{g \choose i},\ 0\le i\le g.$$ This would imply that the Euler characteristic $\chi(\mathcal{O}_X)=\Sigma_{i=0}^g(-1)^i {g\choose i}=0$. This would mean that the arithmetic genus $g_a(X)=(-1)^g(\chi(\mathcal{O}_X)-1)=(-1)^{g+1}$. This obviously aligns with the fact that the genus of an elliptic curve is $1$. That would mean that the genus of an abelian varieties of even dimension is negative.
Is that possible?
Have I made some error in computation?
Your computations are correct. There is no error; in particular, the arithmetic genus can be negative. In fact, it can take on any integer value, see this question.