Yes, for example $(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ the Prüfer $p$-group subfactor.

More precisely:

Let $p$ be a prime number and let the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \mathbb{Z}/p^3\mathbb{Z} \hookrightarrow \cdots$

given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is $\mathbb{Z}(p^{\infty})$ the Prüfer $p$-group.

(Note that $\mathbb{Z}(p^{\infty})$ is the Pontryagin dual of the group of $p$-adic integers $\mathbb{Z}_p$)

$\mathbb{Z}(p^{\infty})$ admits no extra subgroups so its lattice is an infinite homogeneous single chain.

Now it is a discrete group so by Galois correspondence for discrete group (theorem 3.13 p 48 of this paper of Izumi-Longo-Popa) for a given outer action of $\mathbb{Z}(p^{\infty})$ on the hyperfinite ${\rm II}_1$ factor $R$ then the subfactor $(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ admits an infinite homogeneous single chain lattice.

*Remark*: The planar subalgebra of $P(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ generated by all the biprojections "should be" $TLJ_{p^{1/2}}^{* \infty}$. So for any $\delta$ we should have the example $TLJ_{\delta}^{* \infty}$.

More generally, any infinite free composition of a maximal subfactor should be an example.