$$
XXXXIII \\
XXIIXXI \\
XIXIXIX \\
ZZ^\dagger Z^\dagger Z III \\
ZZ^\dagger II Z^\dagger Z I \\
Z I Z^\dagger I Z^\dagger I Z
$$
After I asked the question I played around with the generators by hand and wasn't really getting anywhere, trying to use inverses for both the $ X $ and $ Z $ type stabilizers. But then I was like what if I just take all the $ X $ type stabilizers as is and then just use inverses for the $ Z $ type ones? And that seemed worth trying since treating $ X $ and $ Z $ symmetrically wasn't working out for me. So then I did that and just took all the $ X $ type to be regular with no inverses. Then I looked at $ Z $ and was like ok if $ X $ is all regular then to commute the $ Z $ needs to be half regular half inverses. And without a loss of generality I can probably take the $ Z $ on the first qubit to always be regular. From there it seems like the placement of the two $ Z^\dagger $ is uniquely determined. I don't know if this method would work for qudit CSS codes in general, but maybe?
I guess the idea of using this in a more general context would be if you have a qubit CSS code with $ H_X=H_Z $ then you can take all the qudit stabilizers of one type (say $ X $ type) to be regular. Then for the stabilizers of the other type (say $ Z $ type) you have to take half of the $ Z $ regular and half as $ Z^\dagger $. Then hopefully you can arrange it all in a way so the phases cancel out and all the $ X $ and $ Z $ qudit stabilizer generators commute?
It was definitely not what I expected with the asymmetry of treating $ X $ and $ Z $ differently, especially since the 5 qudit code treats $ X $ and $ Z $ so nice and symmetrically. So overall pretty weird and I guess I'm hoping for another answer that is bette/nicer/more symmetric/ more systematic/has cool theory behind it.
Some properties of this code:
Logical $ X $ is implemented by $ X^{\otimes 7} $. Logical $ Z $ is implemented by $ ZZ^\dagger Z^\dagger Z Z^\dagger ZZ $.
For a qudit of prime dimension $ p $ the $ H $ gate is essentially a Fourier transform over the field with $ p $ elements, so it is essentially canonical.
The effect of conjugating by $ H $ is $ X \mapsto Z $, $ Z \mapsto X^\dagger $ and the effect of conjugating by $ H^\dagger $ is $ X \mapsto Z^\dagger $, $ Z \mapsto X $ . So you can check that $ HH^\dagger H^\dagger HH^\dagger HH $ implements logical $ H $ (that is, it preserves the stabilizer group and it takes logical $ X $ to logical $ Z $ and logical $ Z $ to logical $ X^\dagger $)
The choice of phase gate is not so canonical. There are two pretty reasonable conventions. One is given here What is the set of generators for the qutrit Clifford group? where the phase if $ \omega^{j(j+1)/2} $. A perfectly good alternative convention is choosing the phase to be $ \omega^{j(j-1)/2} $, which is used, for example, in Gottesman's book (this is all in section 8.4 Qudit Clifford Group, Gottesman calls the phase gate $ B $, I do not know why). Gottesman's convention is probably more aesthetically pleasing but for no good reason I am going to stick to the other convention.
The effect of conjugating by $ S $ is $ X \mapsto ZX $ (it would map to $ XZ $ in the alternate convention) and $ Z \mapsto Z $. Similarly, the effect of conjugating by $ S^\dagger $ is $ X \mapsto Z^\dagger X $, $ Z \mapsto Z $ . So you can check that $ SS^\dagger S^\dagger SS^\dagger SS $ implements logical $ S $ (that is, it preserves the stabilizer group and it takes logical $ X $ to logical $ ZX $ and logical $ Z $ to logical $ Z $).
Thus the transversal gate group of the qutrit Steane code is the single qutrit Clifford group, as expected.
Note that the stabilizer generators given here works for any qudit of prime dimension to yield a $ [[7,1,3]]_p $ code. And the transversal gates also work for any prime $ p $. So this $ [[7,1,3]]_p $ code has transversal gate group the single qupit Clifford group for $ p $ prime.
