The real Pauli group is defined as the subgroup of the Pauli matrices with only real entries. In particular, it contains $Y_i Y_j = - X_i Z_i X_j Z_j$. One can show that it is generated by $X_i$ and $Z_j$. The real Clifford group is the orthogonal normalizer of the Pauli group, or equivalently, the subgroup of the Clifford group given by real matrices.
The real Clifford group and its generators are first described in Calderbank, Rains, Shor, and Sloane (1997). These are
- Hadamard gate on the first qubit, $H_1$,
- Linear reversible circuits $|x\rangle \mapsto |Ax\rangle$ for $A\in\mathrm{GL}(n,\mathbb{F}_2)$,
- Diagonal gates $|x\rangle \mapsto (-1)^{q(x)}|x\rangle$ where $q$ is a binary quadratic form.
Finally, to adress your question: Assume $n \geq 2$ qubits and let us consider the generators $H_i$ and $CX_{ij}$ for all $i,j$. Thus by assumption we have (1.) and moreover the CNOT gates $CX_{ij}$ generate the linear reversible circuits (2.). To obtain the diagonal gates, note that every binary quadratic form can be written as $q(x) = \sum_{i<j} x_i R_{ij} x_j$. To realize such a gate, apply $CZ_{ij}$ to every qubit pair for which $R_{ij}=1$. Clearly, $CZ_{ij}= H_j CX_{ij} H_j$ and thus we have shown that we can generate the entire real Clifford group.
If you wonder how we generate Paulis, note that we have $Z_1 = (CX_{12}\, H_2)^4$.
EDIT: I noticed that for $n=1$, this argument does not work. In this case, generators are $H$ and $Z$. Formally, the quadratic form in point 3 needs to be supplemented by a linear form (not sure whether Calderbank et al. meant that in the first place).
