Denote $L_1=L_1[0,1]$ The lattice of closed ideals in $\mathcal{L}(L_1)$ includes the chain $$ \{0\}\subsetneq\mathcal{K}(L_1)\subsetneq\mathcal{FS}(L_1) \subsetneq\mathcal{J}_{\ell_1}(L_1)\subsetneq\mathcal{S}_{L_1}(L_1) \subsetneq\mathcal{L}(L_1). $$ I believe it's still an open question whether the lattice contains any more distinct elements than the above listed.

One strategy for producing many closed ideals very quickly is to use Schreier-Rosenthal operator ideals, and this has worked, for instance, in the cases $\mathcal{L}(\ell_1\oplus\ell_\infty)$. In particular, given a countable ordinal $\xi$, we could seek some special operator $A\in\mathcal{R}_\zeta(L_1)\setminus\mathcal{R}_\xi(L_1)$ for some $\zeta>\xi$.

Here's how it worked for $\mathcal{L}(\ell_1\oplus\ell_\infty)$. Let $T_\xi$ denote the $\xi$-Tsirelson space, and let $I_{1,\xi}:\ell_1\to T_\xi$ be the formal identity. Of course we can find an embedding $J_{\xi,\infty}:T_\xi\to\ell_\infty$. Then $J_{\xi,\infty}\circ I_{1,\xi}$ is class $\mathcal{R}$ but not class $\mathcal{R}_\xi$. By a result of Beanland/Freeman we can find $\zeta>\xi$ such that it is class $\mathcal{R}_\zeta$ as well. Then use transfinite induction.

Adapting this approach for $\mathcal{L}(L_1)$ presents serious difficulties. Of course $L_1$ contains a complemented copy of $\ell_1$, so we can still use a formal identity map from $\ell_1$ onto any space with an unconditional basis. But we can't use $T_\xi$ because it doesn't embed into $L_1$. (We know this because every infinite-dimensional subspace of $L_1$ contains a copy of $\ell_p$ for some $1\leq p\leq 2$.)

But perhaps there is a way around this. For any scalar sequence $(a_n)$ we can define $$\|(a_n)\|_\xi=\|(a_n)\|_2\vee\sup_{F\in\mathcal{S}_\xi}\|(a_n)_{n\in F}\|_1$$ This generates a 1-unconditional basic sequence whose closed linear span we denote $X_\xi$. Is it possible that $X_\xi$ (or something like it) embeds into $L_1$? We would also need to make sure that $\ell_1$ doesn't embed into $X_\xi$, but that seems fairly likely given the latter's construction.

In general, to make this work we need to find a space $X_\xi$ with a basis which is $\xi$-equivalent to $\ell_1$ and embeds into $L_1$, but doesn't contain a copy of $\ell_1$. Frankly, I believe this is asking for too much. Nevertheless, I thought it would be worth seeing what y'all thought before I give up on the idea. And if there is no such space, perhaps that would be worth proving anyway.

Thanks guys!