The fewer qubits it uses the better. Also, it would be great if the code admitted a transversal logical $CNOT$.
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Every CSS code has transversal CNOT gate. Moreover, if the $X$ and $Z$ sectors of the stabilizer are isomorphic then the code has transversal Hadamard. See this answer for proofs of both facts. The CSS construction is also described in section 10.4.2 on page 450 in Nielsen & Chuang.
This leads us to the smallest non-trivial code with transversal CNOT and Hadamard. Namely, the $[\![4,2,2]\!]$ code with stabilizer group generated by $XXXX$ and $ZZZZ$. See this question for arguments that rule out smaller codes.
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$\begingroup$ I believe that CSS codes are also the only codes with transversal CNOT $\endgroup$– unknownJan 2 at 19:03
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$\begingroup$ May you expand with the corresponding logical Hadamard? $\endgroup$ Jan 2 at 21:23
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1$\begingroup$ @unknown It's not hard to show that CSS codes are the only stabilizer codes with transversal CNOT. Is it true that non-additive codes cannot have transversal CNOT? $\endgroup$ Jan 2 at 23:22
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1$\begingroup$ @DanieleCuomo You can apply simultaneous logical Hadamard to both encoded qubits using $HHHH$. However, note that $|00_L\rangle$ has entanglement across every partitioning of the physical qubits while $|{0+}_L\rangle$ lacks entanglement across some partitions, so Hadamard on just one of the two encoded qubits is not transversal. If this is a problem you can ignore the second logical qubit and pretend you only have one. Then $HHHH$ is transversal Hadamard on your only logical qubit. N.B. such ignored qubits can be utilized to reduce weights of measured operators, see subsystem codes. $\endgroup$ Jan 3 at 0:25
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$\begingroup$ @AdamZalcman can I state, as immediate inference from your answer, that: there is no code admitting logical hadamard and at the same time able to detect only X (xor Z) noise? As for the case of the repetition code. $\endgroup$ Jan 4 at 15:52