I think I found a proof that in a general Artin algebra one has that the projective dimension of $D(A)$ is equal to the Gorenstein projective dimension of $D(A)$. The proof is simple in case I made no mistake. But it relies on something which I expected to be wrong so I have not thought about this at the beginning.

A complex $X^{\bullet}=(X^n,d^n)$ is called right GP-acyclic in case $Hom(G,X^{\bullet})$ is acyclic for each Gorenstein projective module $G$.
A Gorenstein projective resolution of a module $M$ is an acyclic complex $....G_2 \rightarrow G_1 \rightarrow G_0 \rightarrow M \rightarrow 0$ that is also right GP-acyclic.

Now a module $M$ has Gorenstein projective dimension at most $n$ iff for every right GP-acyclic complex $0 \rightarrow K \rightarrow G^{n-1} \rightarrow .. \rightarrow G^0 \rightarrow M \rightarrow 0$ with $G^i$ Gorenstein projective has the property that $K$ is Gorenstein projective.

Now I show that a minimal projective resolution $(P_i)$ of an indecomposable injective module $I$ of infinite projective dimension is right GP-acyclic and $\Omega^{i}(I)$ is never Gorenstein projective, which shows that $I$ also has infinite Gorenstein projective dimension. That the minimal projective resolution is GP-acyclic is a direct consequence that $Hom(G,P_i) \rightarrow Hom(G, \Omega^i(I))$ is surjective for each Gorenstien projective module $G$, since $Ext^1(G,\Omega^i(I))=0$ for each $i \geq 0$ (see proposition 2.2. in https://arxiv.org/pdf/1710.03066.pdf ). But also by proposition 2.2. of https://arxiv.org/pdf/1710.03066.pdf , $\Omega^i(I)$ is never Gorenstein projective and thus the Gorenstein projective dimension of $I$ has to be infinite.