Let $P$ be the wedge of all the $X_i$s. Up to homotopy equivalence, $X$ is the homotopy coequalizer of the identity and the shift maps from $P$ to itself. The Milnor exact sequence arises by analyzing the resulting long exact sequence $$ ...\rightarrow h^n(\Sigma P) \xrightarrow{1 - \text{shift}} h^n(\Sigma P) \xrightarrow{d^*} h^n(X) \xrightarrow{p^*} h^n(P) \xrightarrow{1- \text{shift}} h^n(P) \rightarrow \dots,$$ where $P \xrightarrow{p} X \xrightarrow{d} \Sigma P \xrightarrow{1-\text{shift}} \Sigma P$ is a cofibration sequence.
From this, one sees that an element $x \in h^n(X)$ is in $I$ (defined as you did) if and only if $x$ is in the image of $d^*: h^n(\Sigma P) \rightarrow h^n(X)$. Since cup products vanish in $h^*(\Sigma P)$ (the cohomology of a suspension), it follows that $I^2 = 0$, as you suspected.
[Homology groups here are all reduced.]