Is there a maximal subgroup of depth 3?

This isn't a full answer.

First let’s translate this into purely group theoretic language.

The vertex at depth $0$ is the trivial $G$-rep, the vertex at depth $1$ is the trivial $H$-rep, the vertices at depth $0$ or $2$ are the $G$-irreps in $\mathrm{Ind}_H^G 1$, the vertices at depth $1$ or $3$ are the $H$-irreps in $\mathrm{Res}_H\mathrm{Ind}_H^G 1$, the vertices at depth $0$, $2$, or $4$ are the $G$-irreps in $\mathrm{Ind}_H^G \mathrm{Res}_H\mathrm{Ind}_H^G 1$, etc.

Let's first ask which $H$ and $G$ irreps appear eventually after successive induction-restriction. These are exactly the reps which are trivial when restricted to $N =\cap_g gHg^{-1}$. We can replace $H \subset G$ with $H/N \subset G/N$ without changing the graph (or the subfactor). So WLOG assume $N = \{1\}$ and so $G$ is a transitive subgroup of $S_n$ with $n=|G:H|$.

So what you want is a maximal subgroup $H \subset G$ such that $H$ is non-trivial (so there are vertices at depth $3$), and every $G$-irrep occurs in $\mathrm{Ind}_H^G 1$ (so there are no vertices at depth $4$).

If the index $|G:H|$ is a prime $p$. Take $P$ a Sylow $p$-subgroup of $G$. This is just a choice of $p$-cycle in $G$. The double coset space $P\backslash G/ H$ is trivial since $P$ is a transitive subgroup of $S_p$. So $$\mathrm{Res}_H \mathrm{Ind}_P^G 1 = \mathrm{Ind}_1^H 1$$ only has one copy of the trivial. Hence any nontrivial irrep $W$ in $\mathrm{Ind}_P^G 1$ when restricted to $H$ has no trivial subrep. Hence $W$ doesn’t appear in $\mathrm{Ind}_H^G 1$.

Then $\mathrm{Ind}_P^G 1$ is a trivial representation of $G$, and so $P=G$ is a $p$-group. But $G \subset S_p$ and $p$ only divides $p!$ once, hence $G$ is cyclic of order $p$ and $H$ is trivial. It follows that $H \subset G$ is depth $2$, contradiction with depth $3$.

Conclusion: There is no maximal subgroup of depth $3$ and prime index, i.e. $I_3 \cap \mathbb{P} = \emptyset$.

As Noah points out, you are looking for some (core-free) maximal subgroup $H<G$ such that $1_H^G$ has nonzero inner product with every irreducible character.

Say $G=L_2(p)$ with $p$ prime and $p \equiv 1 \bmod 8$. Then $G$ has a maximal subgroup $H \cong S_4$. Every element $h \in H$ has diagonalizable preimage in $SL_2(p)$, and the conjugacy class of $h$ in $G$ is determined by $|h|$. Note $|h| \in \{1,2,3,4\}$ and that $G$ has a unique conjugacy class of elements of each such order.

Let $h_2$, $h_3$ and $h_4$ represent, respectively, the conjugacy classes of elements of orders $2,3,4$. Using the facts above and direct calculation, you should get that for any irreducible character $\chi$ of $G$,

$$ \langle 1_H^G,\chi \rangle=\frac{1}{24}(\chi(1)+9\chi(h_2)+8\chi(h_3)+6\chi(h_4)). $$

Now, assuming $\chi$ is not the trivial character, $\chi(1)$ is at least $\frac{p+1}{2}$. On the other hand, each of $\chi(h_2)$, $\chi(h_3)$ and $\chi(h_4)$ is a sum of at most two roots of unity. Thus, for large enough $p$, you get $\langle 1_H^G,\chi \rangle>0$ for all irreducible $\chi$, and $H$ has depth three.

I found the character table for $L_2(q)$ at the web page of Jeffrey Adams,

http://www.math.umd.edu/~jda/characters/characters.pdf

You might be able to pull a similar trick in other cases - take some group $H$ and a (projective) complex irrep of $H$. This should in many cases reduce to an embedding of $H$ in some simple groups of fixed Lie type over various fields, some of these embeddings making $H$ maximal. For large enough fields, you might get depth three.