Interesting question! Here is a generalization which includes the Dedekind domain case:

>**Proposition:** Let $R$ be a Noetherian regular domain with $n=\dim R$  and $I\subset R$ an ideal such that $R/I$ is artinian and Gorenstein. Then for a finitely generated $R$-module $M$, we have $Ext^n(R/I,M)\cong M/IM$.

(when $R$ is Dedekind $R/I$ is locally a quotient of a DVR, so it is locally a hypersurface, thus Gorenstein).

**Proof**. It is enough to prove for $M=R$ because of the following 

>Claim: if $N$ is an artinian $R$-module then the natural map $Ext^n_R(N,R)\otimes M \to Ext^n_R(N,M)$  is an isomorphism. 

Here's why: We have $pd_RN = n$. Let $P_{\bullet}: 0\to P_n \to \cdots \to P_0$ be a projective resolution of $N$. Then one compute the LHS by taking the cohomology at the end of $Hom(P_{\bullet},R)$ and tensor with $M$, and the RHS by taking cohomology at the end of $Hom(P_{\bullet},M)$. But tensor product is right-exact, proving that we get the same answer. 

Assuming we can prove for $M=R$, then apply the claim with $N=R/I$ completes the proof.

Now we prove it for $M=R$. Since $R/I$ is artinian, we only need to prove it after localizing at each maximal ideal containing $I$. Thus we can assume $(R,m)$ is local. But by Local Duality $Ext^n(R/I,R)$ is isomorphic to the  Matlis dual of $H^0_m(R/I) =R/I$. But since $R/I$ is Gorenstein this dual is actually isomorphic to  $R/I$. QED