Yes, I think the statement is also true for $n >2$ even (see Guntram's answer of the question Automorphisms of $SL_n(\mathbb{Z})$).
As a second reference a paper of O'Meara may serve. It can be found here. The following specializes O'Meara's result to the case $SL_n(\mathbb{Z})$:
- Note that in our context $\mathbb{o} = \mathbb{Z}, K = \mathbb{Q}, M = \mathbb{Z}^n.$.
- Since the only automorphisms of the field $\mathbb{Q}$ is the identity, semi-linear here means linear.
- By 1.13, $RL_n(M)$ is the center of $GL_n(\mathbb{Z})$, i.e. $RL_n(M) = \lbrace I_n, -I_n \rbrace \cong \mathbb{Z}/2\mathbb{Z}$ and since $SL_n(\mathbb{Z})$ is a perfect group, the only homomorphism $SL_n(\mathbb{Z}) \to RL_n(M)$ is the trivial one. Hence the radical automorphisms $P_{\chi}$ in 5.4 Theorem A are just the identity.
- By 5.4 Theorem A it follows that the automorphism of $SL_n(\mathbb{Z})$ are of the form $\Phi_g, \Psi_h$ that are given in matrix form (end of page 87) by $$S \mapsto PSP^{-1}, S \mapsto P^{-1}(S^T)^{-1}P$$ for some $P \in GL_n(\mathbb{Q})$.
- By 5.5 Theorem B, $\Phi_g$ is an automorphism of $SL_n(\mathbb{Z})$ iff $P*\mathbb{Z}^n = q\mathbb{Z}^n$ for some $q \in \mathbb{Q}$ (where $*$ is matrix-vector multiplication). It's not hard to see that this implies $P = qQ$ for some $Q \in GL_n(\mathbb{Z})$. Thus $\Phi_g: S \mapsto QSQ^{-1}$.
- By 5.6 Theorem C, $\Psi_h$ is an automorphism of $SL_n(\mathbb{Z})$ iff $h(M) = qM^\ast$ for some $q \in \mathbb{Q}$, where $M^*$ is the free $\mathbb{Z}$-module whose base is the dual base of the standard vectors in $\mathbb{Q}^n$ (see page 65). If $P$ is the matrix of $h$ in the standard base of $\mathbb{Q}^n$ and its dual base in $(\mathbb{Q}^n)^\ast$ then, again, it's not hard to see that $P = qQ$ for some $Q \in GL_n(\mathbb{Z})$. Thus $\Psi_h: S \mapsto Q^{-1}(S^T)^{-1}Q$.
In particular each automorphism of $SL_n(\mathbb{Z})$ is induced by an automorphism of $GL_n(\mathbb{Z})$ (what was left open in the Hua-Reiner paper).