Not a complete answer (for lack of time, will try to add later). Since $\min(a,b)=(a+b-|a-b|)/2$, your question is really about upper bounding $E|2X-n|$, or, equivalently, $E|X-n/2|$:

$$E\min(X,n-X)=n-2E|X-n/2|.$$

You can upper bound $E|X-n/2|$ using Jensen's inequality:
$E|X-n/2|\le\sqrt{E(X-n/2)^2}$.

The latter, if I'm not mistaken, evaluates to
$$ n\sqrt{
p(1-p)/n+p^2-p+1/4
}
=:F(p).$$