The $ [[15,1,3]] $ triorthogonal code implements transversal $ T $. Since it is a CSS code, two blocks will also have a transversal $ CNOT $ gate. To get a universal gate set all that is required is an implementation of the Hadamard gate $ H $.
The Hadamard gate can be implemented via gate teleportation from the state $$ (I \otimes H) \frac{|00\rangle+|11\rangle}{\sqrt{2}}= \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle-|11\rangle) $$
It is my impression that the standard proposal for a universal gate set is to use the $ [[7,1,3]] $ Steane code to implement all the Clifford gates transversally then use magic state distillation to implement the $ T $ gate on the Steane code.
My extremely primitive understanding of these things is that fault tolerantly preparing stabilizer states is easy but fault tolerantly preparing magic states like $ T | + \rangle = \frac{1}{\sqrt{2}}(|0\rangle + e^{i\pi/4}|1\rangle) $ is extremely hard.
If this is the case then wouldn't it be way easier to prepare the stabilizer state $$ (I \otimes H) \frac{|00\rangle+|11\rangle}{\sqrt{2}}= \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle-|11\rangle) $$ (which has stabilizer generators $ XZ,ZX $) and then use gate teleportation to use this state to implement $ H $ fault tolerantly on the $ [[15,1,3]] $ code and so do universal computation with the $ [[15,1,3]] $ code instead of the Steane code.
To summarize my question: if stabilizer states are easy and magic states are hard then wouldn't it be better to do universal computation with $ [[15,1,3]] $ and $ H $ instead of $ [[7,1,3]] $ and $ T $?
