Martin O's answer is very nice. So in an oriented $2n$-manifold with $n$ odd the mod $2$ self-intersection of any $n$-dimensional mod $2$ homology class is $0$.
Looking for a more geometric explanation of that, or anyway an explanation with no Steenrod operations in sight, I came up with the following (which is also related to John Klein's comment):
Let's assume that the given class is represented by an immersed $n$-manifold $M$. The mod $2$ self-intersection number is then the evaluation on the mod $2$ fundamental class of $M$ of the mod $2$ Euler class of the normal bundle of the immersion. So it comes down to the following:
Claim: Let $n$ be odd and suppose that $M$ is a closed $n$-dimensional manifold and $E$ is a rank $n$ vector bundle such that the total space of $E$, considered as a $2n$-manifold, is orientable. Then the mod $2$ Euler class of $E$ is $0\in H^n(M;\mathbb Z/2)$.
Proof: A rank $n$ vector bundle has a twisted integral Euler class, which belongs to $H^n(M;\Gamma)$, where $\Gamma$ is the coefficient system (locally isomorphic to $\mathbb Z$) associated with $w_1(E)$, the obstruction to orientability of $E$. The mod $2$ Euler class is the mod $2$ reduction of this, so it suffices if this twisted integral class is $0$. The (twisted) integral Euler class of a vector bundle of odd rank is always killed by $2$ (this is a standard fact in the oriented case, and it seems clear in the twisted case, too), so it suffices if the group $H^n(M;\Gamma)$ is torsion-free. But by Poincare duality it is isomorphic to $H_0(M;\mathbb Z)$, since $E$ and the tangent bundle of $M$ have the same orientability obstruction by hypothesis.