I worked out a special case of this, so might as well post. Do check out this answer by Torsten Schöneberg for more.
I want to look at the variant of $L=\mathfrak{o}(8)$, where the positive root algebras appear in the upper triangular part. To that end I specify
$$
J=\left(\begin{array}{cccccccc}0&0&0&0&0&0&0&1\\
0&0&0&0&0&0&1&0\\
0&0&0&0&0&1&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&1&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&1&0&0&0&0&0&0\\
1&0&0&0&0&0&0&0
\end{array}\right),
$$
when I get a simple Lie algebra of type $D_4$ via
$$
L=\{x\in \mathfrak{gl}(8)\mid xJ+Jx^t=0\}
$$
basically specifying the space of complex $8\times8$-matrices antisymmetric with
respect to the wrong diagonal.
If $e_{ij}$ stands for the basic matrix with a single $1$ at position $(i,j)$ and zeros elsewhere, then the strictly upper triangular elements of $L$ are spanned by the twelve root vectors
$$x_{12},x_{13},x_{14},x_{15},x_{16},x_{17},x_{23},x_{24},x_{25},x_{26},x_{34},x_{35},$$
where $x_{ij}$ stands for the "antisymmetric pair": $x_{ij}=e_{ij}-e_{9-j,9-i}$,
observe the pair of subscripts in the latter term.
The next task is to match these with the positive roots of $D_4$. The upper left $4\times 4$ block (together with its mirror images in the lower right) forms a copy of $\mathfrak{sl}(4)$, suggesting the placement of some roots. Anyway, we recall that forming commutators corresponds to addition of roots and the roots of the basis $\Delta$ need to be indecomposable (=not sums of two other positive roots). Together with the more basic recollection that commutators of strictly upper triangular matrices take us further away from the main diagonal, this leads to the desire to pack the simple roots close to the main diagonal as possible. It is easy to verify the following scheme works (using the mirror symmetry I only show the part above both diagonals:
$$
\left(\begin{array}{cccccccc}
0&\alpha_1&\alpha_1+\alpha_2&\alpha_1+\alpha_2+\alpha_3&\alpha_1+\alpha_2+\alpha_4&\alpha_1+\alpha_2+\alpha_3+\alpha_4&\alpha_1+2\alpha_2+\alpha_3+\alpha_4&0\\
&0&\alpha_2&\alpha_2+\alpha_3&\alpha_2+\alpha_4&\alpha_2+\alpha_3+\alpha_4&0&\\
&&0&\alpha_3&\alpha_4&0&&\\
\end{array}\right).
$$
Indeed, $x_{12},x_{13},x_{14},x_{23},x_{24},x_{34}$ obey the commutator relations of the corresponding basis elements of $\mathfrak{sl}(4)$ accounting for the linear combinations of $\alpha_1,\alpha_2,\alpha_3$. Then with $x_{35}$ we get the rest
from
$$
\begin{aligned}[]
[x_{23},x_{35}]&=x_{25},\\
[x_{12},x_{25}]&=x_{15},\\
[x_{34},x_{25}]&=x_{26},\\
[x_{12},x_{26}]&=x_{16},\\
[x_{23},x_{16}]&=x_{17}.
\end{aligned}
$$
Observe that the simple root $\alpha_2$ corresponds to the center node in the Dynkin diagram. This is apparent from the list of height two roots.
The theorem in section 14.2. of Humphreys then promises that we can extend the order three graph automorphism $\alpha_2\mapsto \alpha_2, \alpha_1\mapsto \alpha_3\mapsto \alpha_4\mapsto \alpha_1$ into an automorphism $\tau$ of Lie algebras when the images of chosen elements of the root spaces coming from simple roots are prescribed accordingly:
$$
\tau:x_{23}\mapsto x_{23}, x_{12}\mapsto x_{34}\mapsto x_{35}\mapsto x_{12}.
$$
We continue by observing the commutators as follows (step-by-step away from the main diagonal, using the previously computed $\tau(x_{ij})$):
$$
\begin{aligned}
x_{13}=[x_{12},x_{23}]&\mapsto[\tau(x_{12}),\tau(x_{23})]=[x_{34},x_{23}]=-x_{24},\\
x_{24}=[x_{23},x_{34}]&\mapsto[\tau(x_{23}),\tau(x_{34})]=[x_{23},x_{35}]=x_{25},\\
x_{14}=[x_{12},x_{24}]&\mapsto[\tau(x_{12}),\tau(x_{24})]=[x_{34},x_{25}]=x_{26},\\
x_{25}=[x_{23},x_{35}]&\mapsto[\tau(x_{23}),\tau(x_{35})]=[x_{23},x_{12}]=-x_{13},\\
x_{15}=[x_{12},x_{25}]&\mapsto[\tau(x_{12}),\tau(x_{25})]=[x_{34},-x_{13}]=x_{14},\\
x_{26}=[x_{34},x_{25}]&\mapsto[\tau(x_{34}),\tau(x_{25})]=[x_{35},-x_{13}]=x_{15},\\
x_{16}=[x_{12},x_{26}]&\mapsto[\tau(x_{12}),\tau(x_{26})]=[x_{34},x_{15}]=x_{16},\\
x_{17}=[x_{23},x_{16}]&\mapsto[\tau(x_{23}),\tau(x_{16})]=[x_{23},x_{16}]=x_{17}.
\end{aligned}
$$
As further check observe that the root subspaces of a fixed height are necessarily permuted, as expected!
From this we immediate see that within the positive root side the space of fixed points of $\tau$ is spanned by the set
$$
\mathcal{B^+}:=\{x_{23},t_{12}:=x_{12}+x_{34}+x_{35},t_{13}:=x_{13}-x_{24}-x_{25},t_{14}:=x_{14}+x_{15}+x_{26}, x_{16},x_{17}\}.
$$
I have only checked these "positive fixed points" but there were no surprises within the commutators. In other words, the set of fixed points looks like a subalgebra to me. For example,
$$
\begin{aligned}[]
[t_{12},t_{13}]&=-2t_{14},\\
[t_{12},t_{14}]&=3x_{16},\\
[t_{12},x_{23}]&=t_{13}.
\end{aligned}
$$
The recipe in Humphreys, section 14.2, tells us that we only need to use the corresponding linear map on $H$, and the negative root spaces then come out as expected. I carried out all the calculations (aided by Mathematica), and as expected, the transposes of the above matrices give the negative part, and the obvious commutators span the toral part of the fixed points of $\tau$.
After a bit of testing it turns out that we want to single out the following spanning set of $H\cap \mathrm{Inv}(\tau)$:
$$
\begin{aligned}
h_\alpha&:=[x_{23},x_{23}^t]=diag(0,1,-1,0,*),\\
h_\beta&:=[t_{12},t_{12}^t]=diag(1,-1,2,0,*),
\end{aligned}
$$
where the last entry $*$ is a shorthand for the antisymmetric reflection about the wrong diagonal (so the entries repeat with opposite signs and reverse order).
The symmetry of $\tau$ (or the Dynkin diagram) then dictates that the elements of $\mathcal{B}^+$ are all eigenvectors of both $\mathrm{ad} h_\alpha$ as well as $\mathrm{ad} h_\beta$. The corresponding lists of eigenvalues are
$$
\begin{aligned}
\mathrm{ad}(h_\alpha)&=diag(2,-1,1,0,-1,1),\\
\mathrm{ad}(h_\beta)&=diag(-3,2,-1,1,3,0).
\end{aligned}
$$
This is sufficient data to calculate the Killing form. For simplicity I use only the positive root spaces, so the restriction of the true Killing form to $H=\langle h_\alpha, h_\beta\rangle$ would be twice to what I got.
But for the purposes of identifying the resulting root system, the scale is irrelevant. We get as the trace of the composition of adjoint actions:
$$
\kappa(h_\alpha,h_\alpha)=8, \kappa(h_\alpha,h_\beta)=-12,
\kappa(h_\beta,h_\beta)=24.
$$
From this we get the metric data of the root system, first the duals of roots $t_\alpha=h_\alpha/4$, $t_\beta=h_\beta/12$, and then the data Torsten Schöneberg "spoiled": $||\alpha||=\sqrt3 ||\beta||$ and the angle
in-between $=5\pi/6$. Do ask, if you want more details about converting the values of the Killing form into the metric data on roots!
In the order I listed (see the lists of entries in
$\mathrm{ad}(h_\alpha)$ and $\mathrm{ad}(h_\beta)$), we see that the roots in the listed order are
$$
\alpha,\beta,\alpha+\beta,\alpha+2\beta,\alpha+3\beta,2\alpha+3\beta.
$$
$G_2$ it is!
In the comment exchange below I was worried about the Lie algebra of type $G_2$ not having an irreducible representation of dimension $8$. Behold!
Let $c(1),\ldots,c(14)$ be the coefficients of the basis
$$
h_\alpha, h_\beta, x_{23},t_{12},t_{13},t_{14},x_{16},x_{17},
x_{23}^t,t_{12}^t,t_{13}^t,t_{14}^t,x_{16}^t,x_{17}^t
$$
in this order. Then the generic matrix of this algebra looks like
$$
\left(
\begin{array}{cccccccc}
c(2) & c(4) & c(5) & c(6) & c(6) & c(7) & c(8) & 0 \\
c(10) & c(1)-c(2) & c(3) & -c(5) & -c(5) & c(6) & 0 & -c(8) \\
c(11) & c(9) & 2 c(2)-c(1) & c(4) & c(4) & 0 & -c(6) & -c(7) \\
c(12) & -c(11) & c(10) & 0 & 0 & -c(4) & c(5) & -c(6) \\
c(12) & -c(11) & c(10) & 0 & 0 & -c(4) & c(5) & -c(6) \\
c(13) & c(12) & 0 & -c(10) & -c(10) & c(1)-2 c(2) & -c(3) & -c(5) \\
c(14) & 0 & -c(12) & c(11) & c(11) & -c(9) & c(2)-c(1) & -c(4) \\
0 & -c(14) & -c(13) & -c(12) & -c(12) & -c(11) & -c(10) & -c(2) \\
\end{array}
\right)
$$
It won't take you more than a few seconds to spot what's going on! The fourth anf the fifth columns (as well as rows) are each others duplicates. Meaning that the subspace spanned by the column vector $(0,0,0,1,-1,0,0,0)^t$ is annihilated by the entire Lie algebra. Therefore the given 8-dimensional rep has a 1-dimensional trivial subrep. The 7-dimensional complement is the smallest faithful irreducible rep of $G_2$.
