I don't know what you can say in general but there is a nice family of examples given by decomposing into blocks.
I'll work over $\mathfrak{su}(n)$ as you can just add the centre back in to $\mathfrak{g}^\perp$. Consider the (orthogonal) decomposition of $\mathfrak{su}(2n) = \mathfrak{s}\oplus \mathfrak{t} \oplus \mathfrak{m}$ where
$$ \mathfrak{s} = \left\{\left.\begin{pmatrix}aI & bI \\ cI & dI\end{pmatrix}\right|\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \mathfrak{su}(2)\right\} \cong \mathfrak{su}(2)$$
$$ \mathfrak{t} = \left\{\left.\begin{pmatrix}A & 0 \\ 0 & A\end{pmatrix}\right|A\in \mathfrak{su}(n)\right\}\cong \mathfrak{su}(n)$$
$$ \mathfrak{m} = \mathfrak{su}(2n) \cap \left\{\left.\begin{pmatrix}A & B \\ C & -A\end{pmatrix}\right|A,B,C\in \mathfrak{sl}(n, \mathbb{C})\right\}$$
Then $\mathfrak{m}$ is isomorphic to $\mathfrak{s}\otimes \mathfrak{t}$ as a $\mathfrak{s}\oplus \mathfrak{t}$-module. Thus you can take $\mathfrak{g} = \mathfrak{s}$ to get a decomposition of $\mathfrak{g}^\perp \cong \mathfrak{s}^{n^2-1} \oplus \mathfrak{t}$ or take $\mathfrak{g} = \mathfrak{t}$ to get $\mathfrak{g}^\perp \cong \mathfrak{t}^{3} \oplus \mathfrak{s}$.
Note for context: I know this decomposition from the non-compact and complex setting but I think it still works in this compact setting. More generally, this comes from a family of decompositions of the form $\mathfrak{sl}_2 \oplus \mathfrak{t} \oplus (S \otimes T)$ where $\mathfrak{sl}_2$, $\mathfrak{t}$ commute, $S,T$ are $\mathfrak{sl}_2$, $\mathfrak{t}$ modules respectively and $S$ is isomorphic to $\mathfrak{sl}_2$ ($T$ is not always isomorphic to $\mathfrak{t}$). These exist in $\mathfrak{sl}_{2n}, \mathfrak{sp}_{2n}, \mathfrak{so}_{n}, \mathfrak{so}_{4n}$ (different form in the latter) and $\mathfrak{e}_7$. Thus you should be able to construct more examples by taking the compact real form of each of these embedded inside some $\mathfrak{su}(n)$. Even more generally, I think you can construct examples similar to this by choosing a copy of $\mathfrak{sl}_2$ sitting across a set of "root $\mathfrak{sl}_2$'s" i.e. $\mathfrak{s}^\alpha := \mathfrak{g}^\alpha \oplus \mathfrak{g}^{-\alpha} \oplus [\mathfrak{g}^\alpha, \mathfrak{g}^{-\alpha}]$ and $\mathfrak{sl}_2 \cong \mathfrak{g} \leq \bigoplus \mathfrak{s}^\alpha$ (you may need the set of roots to be strongly orthogonal).
