Since this question is still unanswered I thought I might write down some of what you can get out of Baumgartner's paper.
In Baumgartner's notation (see the beginning of section 2), $A(\kappa,\lambda,\mu,\nu)$ means that there exists a family of sets $F$ such that
- $F\subseteq P(\kappa)$,
- $|F| = \lambda$,
- $|X| = \mu$ for all $X\in F$, and
- $|X\cap Y| < \nu$ for all $X,Y\in F$ with $X\neq Y$.
Hence the connection is that $\lambda$ is intersecting (in your notation) if and only if $A(\kappa,\kappa^+,\kappa,\lambda)$ holds.
In Theorem 3.4(a) Baumgartner proves that, assuming GCH, for any cardinals $\nu \le \mu \le \kappa$, $A(\kappa,\kappa^+,\mu,\nu)$ holds if and only if $\mu = \nu$ and $cf(\mu) = cf(\kappa)$. Since we're only interested in the case where $\mu = \kappa$, this implies that, under GCH, $i(\kappa) = \kappa$ for all $\kappa$. Note that this conclusion already follows from bof's comments.
The other side is partly covered by Theorem 6.1, which says: assuming GCH holds in $V$, for any cardinals $\nu \le \kappa \le \lambda$ such that $\nu$ is regular, there is a forcing extension $V[G]$ which preserves the cofinalities (hence cardinals) of $V$, in which $A(\kappa,\lambda,\kappa,\nu)$ is true. Hence you can make $i(\kappa) = \omega$ true for any particular $\kappa$, starting from a model of GCH.
It remains to show the consistency of the statement in your question, i.e. for all $\kappa$ there is some $\alpha \ge \kappa$ such that $i(\alpha) < \alpha$. Maybe someone who knows about class forcing can step in.