Re-Amended Answer:
My guess for $\mathrm{SO}(5)$ appears to have been correct in one sense, but not in another. It's true that a nonzero Weyl tensor in this case has to have stabilizer of dimension at most $4$, but there are two distinct candidates with this property.
First, there is the obvious one, which is the Weyl curvature of $\mathbb{CP}^2\times \mathbb{R}$, which has $\mathrm{U}(2)\subset \mathrm{SO}(4)\subset\mathrm{SO(5)}$ as its stabilizer. However, it turns out that $H = \mathrm{SO}(2)\times\mathrm{SO(3)}\subset \mathrm{SO}(5)$ (which is also of dimension $4$) fixes a nonzero Weyl curvature as well, and this yields a completely different $6$-dimensional orbit of $\mathrm{SO(5)}$ in the vector space of Weyl curvatures. Up to multiples, these two are the only $6$-dimensional orbits.
For $n=5$, there are no other nontrivial orbits of dimension $6$ or less. There is a $7$-dimensional orbit, which is the Weyl curvature of the $5$-dimensional symmetric space $\mathrm{SU}(3)/\mathrm{SO}(3)$. Its stabilizer is the subgroup of $\mathrm{SO}(5)$ that is the irreducibly-acting subgroup isomorphic to the holonomy of this space, i.e., $\mathrm{SO}(3)$. Up to multiples, there are no other orbits of dimension $7$ or less than what I have listed.
When you 'bootstrap' the $\mathrm{SO}(2)\times\mathrm{SO(3)}$-stabilizer example to dimensions higher than $5$, you get an element with stabilizer $\mathrm{SO}(2)\times\mathrm{SO(3)}\times\mathrm{SO}(n{-}5)$, which, when $n>5$, has lower dimension than $\mathrm{U}(2)\times\mathrm{SO}(n{-}4)$, so this gives a higher dimensional orbit when $n>5$ than the Weyl curvature of $\mathbb{CP}^2\times\mathbb{R}^{n-4}$.
In low dimensions (but higher than $n=5$) the Weyl curvature of $\mathrm{CP}^2\times\mathbb{R}^{n-4}$ is not optimal. When $n=2m$, there is the Weyl curvature of $\mathbb{CP}^m$, which has $\mathrm{U}(m)\subset\mathrm{SO}(2m)$ as stabilizer, and this group is bigger than $\mathrm{U}(2)\times\mathrm{SO}(2m{-}4)$ for $2 < m < 7$. However, as soon as $m>7$, the group $\mathrm{U}(m)$ has dimension less than the dimension of $\mathrm{U}(2)\times\mathrm{SO}(2m{-}4)$.
Last night, I thought that Weyl curvature of $\mathrm{CP}^2\times\mathbb{R}^{n-4}$ wins for large enough $n$, but this morning, I had another idea and realized that there is an even better candidate in the 'stable' range: I turns out that, for $n\ge 5$, under the subgroup $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)\subset \mathrm{SO}(n)$, the space of Weyl tensors in dimension $n$ has a trivial summand, and hence there is a nonzero Weyl tensor (unique up to multiples) whose stabilizer contains $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$. Since there is no connected Lie group between this subgroup and $\mathrm{SO}(n)$, it follows that the identity component of the stabilizer of this tensor is $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$. Thus, this orbit has dimension $2n{-}4$. I now think that, for sufficiently large $n$ (maybe even $n\ge 9$), this might be the lowest dimensional orbit. It might not be unique, though. For example, when $n=8$, there is an orbit of type $\mathrm{SO}(8)/\mathrm{U}(4)$ and one of type $\mathrm{SO}(8)/\bigl(\mathrm{SO}(2)\times\mathrm{SO}(6)\bigr)$. Both of these orbits have dimension $12$. (NB: Even though these two spaces are isomorphic because of triality, the two Weyl orbits are, of course, quite different.)
There aren't many subgroups of $\mathrm{SO}(n)$ that are larger than $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$ when $n$ is sufficiently large. The one obvious exception, $\mathrm{SO}(n{-}1)$, does not fix a Weyl tensor. That's why I'm thinking that this one will win for sufficiently large $n$.
I haven't thought seriously about $\mathrm{SO}(1,n{-}1)$ yet, but, probably, there are at least two orbits of minimal dimension, namely $\mathrm{SO}(1,n{-}1)/\bigl(\mathrm{SO}(1,1)\times\mathrm{SO}(n{-}2)\bigr)$ and $\mathrm{SO}(1,n{-}1)/\bigl(\mathrm{SO}(1,n{-}3)\times\mathrm{SO}(2)\bigr)$, and there might be another 'degenerate' one that fills in the gap between these two.
