Is the dualizing sheaf on a Cohen-Macaulay scheme reflexive?
The answer is yes. Let me briefly explain why.
There are several ways to define a canonical sheaf on a projective variety $X$. One of them, that works for any normal $X$ is the following. Let $U$ be the smooth locus of $X$ and $i \colon U \hookrightarrow X$ be the natural inclusion. Then define $$\omega_X:=i_*(\Omega^n_U),$$ where $n:=\dim X$. As the sections of $\Omega^n_U$ do not depend on a subset of codimension $2$, neither do the sections of $\omega_X$. In other words, $\omega_X$ is a so-called normal sheaf. Moreover, by construction $\omega_X$ is also torsion-free, so it is reflexive by the second characterization in [1, Proposition 1.6].
When $X$ is Cohen-Macaulay, this sheaf coincides with the dualizing sheaf in the sense of Serre duality. A different way to define a canonical sheaf for embedded varieties $X \subset \mathbb{P}^{N}$ is to consider $$\omega_X' := h^{-n}(\omega_X^{\bullet}),$$
where $\omega_X^{\bullet}$ is the so-called dualizing complex. Again, when $X$ is Cohen-Macaulay this construction provides a reflexive sheaf, that coincides with $\omega_X$ outside a subset of codimension $\geq 2$. Then reflexivity implies $$\omega_X'\simeq \omega_X,$$
see [2, Section 5].
References.
[1]$\,$ Robin Hartshorne, Stable reflexive sheaves, Math. Ann. 254 (1980), no. 2, 121--176.
[2] $\,$ Sándor J. Kovács, Singularities of stable varieties, Handbook of moduli. Vol. II 159--203.
Just for the record, this does not need CM. The following is true:
Let $Z$ be an excellent scheme that admits a dualizing complex. Then $ω_Z$ is torsion-free and $S_2$ on $Z$. If in addition Z is irreducible, then $ω_Z$ is a reflexive $\mathscr O_Z$ -module.
If there is interest in this, I will add a proof.
As requested, here is a proof: (The following is Lemma 3.7.5 in this paper.)
Lemma Let $Z$ be an excellent scheme that admits a dualizing complex. Then $\omega_Z$ is torsion-free and $S_2$ on $Z$. If in addition $Z$ is normal, then $\omega_Z$ is a reflexive $\mathscr O_Z$-module.
Proof. The statement is local, so we may assume that $Z$ is a noetherian affine local scheme. Then, since it admits a dualizing complex, it can be embedded into a finite dimensional Gorenstein affine local scheme W as a closed subscheme by [Kaw02, Cor. 1.4]. Being Gorenstein and local, $W$ must be pure dimensional. Let $r = \mathrm{codim}(Z, W)$. Then by [Mat80, (16.B) Theorem 31(i)] there exists a length $r$ regular sequence in the ideal of $Z$ in $W$ and let $W'$ be the common zero locus. Then $W'$ is also Gorenstein, $Z\subseteq W$ is a closed subscheme and $\dim Z = \dim W$ .
It follows that $\omega_ {W'}^\bullet \simeq \omega_{W'}[m]$, where $\omega_{W'}$ is a line bundle on ${W'}$. By Grothendieck duality $\omega^\bullet_ Z\simeq RHom_{W'}(\mathscr O_Z, \omega^\bullet_{W'})$, and since $\dim Z=\dim W'$, $\omega_Z\simeq Hom_{W'}(\mathscr O_Z, \omega_{W'})$ hence it is indeed torsion-free on $Z$.
By [Stacks Project, Tag 0AWE] $\omega_Z$ is $S_2$. In particular, if $Z$ is normal, then it is reflexive by [Stacks Project,Tag 0AVB].
As far as non-trivial examples of schemes admitting dualizing complexes are concerned, try anything that can be embedded (locally) into a Gorenstein scheme. In fact, those are exactly the ones you are looking for.