Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound $n/2$ on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.
So, $$I_n^*=n/2$$ for even $n$.
It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.
So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.
Conjecture: $I_n^*=n/2$ for all natural $n\ge2$.
Remark 1: Obviously, $I_1^*=0$. In view of the above, it is enough to show that $I_n^*=n/2$ for all odd $n\ge3$. Moreover, in view of the above cancellations, it is enough to show that $I_3^*=3/2$.