How to create the logical $|0_Lrangle$ state for the Steane's 7-qubit code?

Here is a circuit that can create the desired state (similar ideas were discussed in this answer), if all mentioned measurements yield $|0rangle$ state:

or in a more compact form (the circuits are constructed via quirk). The first three qubits are ancillary qubits and the rest are the qubits where $|0_Lrangle$ will be created if after the measurements all ancillary qubits are in the $|000rangle$ state, otherwise one should repeat the procedure until the desired measurement result will be archived.

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Let's go step by step. The circuit has three parts and in each step, we assume that the measured state of the ancillary is $|0rangle$. The normalization factors are omitted.

1. The state after measuring the first ancillary qubit in the $|0rangle$ state: (measuring $XIXIXIX$ stabilizer)

$$|000rangle big( |0000000rangle + |1010101rangle big)$$

2. The state after measuring the second ancillary qubit in the $|0rangle$ state: (measuring $IXXIIXX$ stabilizer)

$$|000rangle big( |0000000rangle + |1010101rangle + |0110011rangle + |1100110ranglebig)$$

3. The state after measuring the third ancillary qubit in the $|0rangle$ state: (measuring $IIIXXXX$ stabilizer)

$$|000rangle |0_Lrangle = |000rangle big( |0000000rangle + |1010101rangle + |0110011rangle + |1100110rangle |0001111rangle + |1011010rangle +|0111100rangle + |1101001rangle big)$$

After disregarding the ancillary qubits that are in the $|000rangle$ state we will have the desired $|0_Lrangle$ state. The probability that each measurement outcome will be $|0rangle$ is $0.5$, thus the probability of creating the $|0_Lrangle$ state with this circuit is $0.125$. For Qiskit, one can implement the circuit mentioned above, then apply any algorithm or gate that is needed on the $|0_Lrangle$ state, and after the computation disregard all the results where ancillary qubits are not in the $|000rangle$ state.