You didn't want examples, but that's what I'm gonna give, for I haven't been doing research related to Lie algebras for ages, and wouldn't know what buzzwords to use to search the publications.
As examples I proffer pairs of Cartan algebras that behave kinda extremally in the sense of your question. This was inspired by the comment Callum made.
So consider the simple Lie algebra $L=\mathfrak{so}_m(\Bbb{C})$ where $m=2\ell+1$ is an odd integer, $\ell\ge2$. I define it as the Lie algebra of antisymmetric $m\times m$ matrices. Let $E_{ij}$ be the $m\times m$ matrix with its sole non-zero entry a $1$ at position $(i,j)$. Let $S_{ij}:=E_{ij}-E_{ji}$ be its antisymmetric
counterpart. The matrices $S_{ij},1\le i<j\le m$ span $L$, and somewhat redundantly
we also have the matrices $S_{ji}=-S_{ij}$.
It is easy to verify that the elements $S_{ij}$ are all $\mathrm{ad}$-semisimple.
As the matrices $S_{12}$, $S_{34}$, $S_{56}$, $\ldots$, $S_{2\ell-1;2\ell}$
commute, and there are $\ell$ of them, they span a Cartan subalgebra $H_1$.
Another Cartan subalgebra $H_2$ is spanned by $S_{23}$, $S_{45}$, $S_{67}$, $\ldots$,
$S_{2\ell;2\ell+1}$. Observe that $H_1$ and $H_2$ intersect trivially.
A third Cartan subalgebra $H_3$ I will use is almost the same as $H_2$. I just
replace the first generating matrix $S_{23}$ with $S_{12}$.
So the subspaces $H_2$ and $H_3$ have a codimension one intersection.
On with the subalgebras the pairs generate. A tool I will use is that whenever $i,j,k$ are three distinct indices, then $S_{ij},S_{jk}$ and $S_{ki}$ span a copy
of the Lie algebra $\mathfrak{so}_3(\Bbb{C})$, and we have the familiar commutator
relations
$$[S_{ij},S_{jk}]=S_{ik},\quad [S_{jk},S_{ki}]=S_{ji},\quad [S_{ki},S_{ij}]=S_{kj}.$$
In other words, we get the third (up to sign) as the commutator of the other two of any such triple.
- The smallest Lie algebra $\mathfrak{A}(H_1,H_2)$ containing both $H_1$ and $H_2$
is then actually all of $L$. This follows easily from the above observation. We get
$S_{13}$ as the commutator of $S_{12}\in H_1$ and $S_{23}\in H_2$. Similarly for the other matrices $S_{i;i+2}$, $i=1,2,\ldots,m-2$. By induction on $d$ we then similarly get all the other matrices of the form $S_{i;i+d}$ with a difference $d$
in the indices as commutators of matrices already known to be in $\mathfrak{A}(H_1,H_2)$.
- On the other hand $\mathfrak{A}(H_2,H_3)$ is much smaller. We easily see that all we need to add to the subspace sum $H_2+H_3$ is the commutator $S_{13}=[S_{12},S_{23}]$. In other words $\mathfrak{A}(H_2,H_3)$ is a direct sum of the algebra $\langle S_{12},S_{23},S_{13}\rangle\simeq \mathfrak{so}_3(\Bbb{C})$ and a toral subalgebra spanned by $S_{45},S_{67},\ldots,S_{2\ell;2\ell+1}$.
It is easy to imagine that something similar might happen in the general case also. The intersection $H_1\cap H_2$ becomes uninteresting, but the rest generate something else. It would not suprise me if something like the above Levi decomposition above would always come out, but I'm too ignorant to be sure.
