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Something is wrong:

If $R$ is a Gorenstein domain of dimension n = 2,3,4, then finite reflexive modules are projective.

Proof: Since both are local properties, we assume $(R,m)$ is local. As Greg pointed out, $\operatorname{Ext}_R^1(M,R) = \operatorname{Ext}^2_R(M,R) = 0$. By local duality, this says $H^{n-1}_m(M) = H^{n-2}_m(M) = 0$. We need to show that $\operatorname{Ext}^i_R(M,R)$ vanish for all $i>2$, iff the lower local cohomologies $H^{n-i}_m(M)$ vanish. But this is clear since $R$ has depth $\ge 2$, any $\operatorname{Hom}_R(N,R)$ will have depth $\ge 2$ (see https://stacks.math.columbia.edu/tag/0AV5), in particular $M$ being reflexive has depth $\ge 2$.

Generalizing your counter-example a bit:

Over a noetherian normal domain, being a reflexive is the same as being a second syzygy (by Auslander). Take any second syzygy of a finite length module, it will have depth 2 - not enough to be depth of the ring in general.

It's super interesting to see your counter-example, especially when the following is true:

If $R$ is a Gorenstein domain of dimension 4, then finite reflexive modules are projective.

Proof: Since both are local properties, we assume $(R,m)$ is local. As Greg pointed out, $\operatorname{Ext}_R^1(M,R) = \operatorname{Ext}^2_R(M,R) = 0$. By local duality, this says $H^3_m(M) = H^2_m(M) = 0$. We need to show that $\operatorname{Ext}^i_R(M,R)$ vanish for all $i = 3,4 $, iff the lower local cohomologies vanish, i.e depth $M \ge 2$. But this is clear since $R$ has depth $4\ge 2$, any $\operatorname{Hom}_R(N,R)$ will have depth $\ge 2$ (see https://stacks.math.columbia.edu/tag/0AV5).

Something is wrong:

If $R$ is a Gorenstein domain of dimension n = 2,3,4, then finite reflexive modules are projective.

Proof: Since both are local properties, we assume $(R,m)$ is local. As Greg pointed out, $\operatorname{Ext}_R^1(M,R) = \operatorname{Ext}^2_R(M,R) = 0$. By local duality, this says $H^{n-1}_m(M) = H^{n-2}_m(M) = 0$. We need to show that $\operatorname{Ext}^i_R(M,R)$ vanish for all $i>2$, iff the lower local cohomologies $H^{n-i}_m(M)$ vanish. But this is clear since $R$ has depth $\ge 2$, any $\operatorname{Hom}_R(N,R)$ will have depth $\ge 2$ (see https://stacks.math.columbia.edu/tag/0AV5), in particular $M$ being reflexive has depth $\ge 2$.

It's super interesting to see your counter-example, especially when the following is true:

If $R$ is a Gorenstein domain of dimension 4, then finite reflexive modules are projective.

Proof: Since both are local properties, we assume $(R,m)$ is local. As Greg pointed out, $\operatorname{Ext}_R^1(M,R) = \operatorname{Ext}^2_R(M,R) = 0$. By local duality, this says $H^3_m(M) = H^2_m(M) = 0$. We need to show that $Ext^i_R(M,R)$ vanish for all $i = 3,4 $, iff the lower local cohomologies vanish, i.e depth $M \ge 2$. But this is clear since $R$ has depth $4\ge 2$, any $\operatorname{Hom}_R(N,R)$ will have depth $\ge 2$ (see https://stacks.math.columbia.edu/tag/0AV5).

It's super interesting to see your counter-example, especially when the following is true:

If $R$ is a Gorenstein domain of dimension 4, then finite reflexive modules are projective.

Proof: Since both are local properties, we assume $(R,m)$ is local. As Greg pointed out, $\operatorname{Ext}_R^1(M,R) = \operatorname{Ext}^2_R(M,R) = 0$. By local duality, this says $H^3_m(M) = H^2_m(M) = 0$. We need to show that $\operatorname{Ext}^i_R(M,R)$ vanish for all $i = 3,4 $, iff the lower local cohomologies vanish, i.e depth $M \ge 2$. But this is clear since $R$ has depth $4\ge 2$, any $\operatorname{Hom}_R(N,R)$ will have depth $\ge 2$ (see https://stacks.math.columbia.edu/tag/0AV5).