Consider the short exact sequence $$0 \to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0.$$ Note that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q) = 0$ for all $i$: it is torsion since $N$ is torsion, but multiplication by $n \in \mathbb Z_{>0}$ is an isomorphism since it is so on $\mathbb Q$. Thus, the sequence above induces isomorphisms $$\operatorname{Ext}^{i-1}_\mathcal C(N,\mathbb Q/\mathbb Z) \stackrel\sim\to \operatorname{Ext}^i_\mathcal C(N,\mathbb Z).$$ Thus, we only have to show that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z) = H^i(G, N^D)$. I will treat the finite group case, and trust that the OP can carry out the limit argument to treat the profinite group $\hat {\mathbb Z}$ (assuming that you are working with discrete [topological] $G$-modules).

Note that if $P$ is a projective $\mathbb Z[G]$-module, then $\operatorname{Hom}_\mathbb Z(P,\mathbb Q/\mathbb Z)$ is an acyclic $\mathbb Z[G]$-module. Indeed, we prove this in three steps:

1. Let $P = \mathbb Z[G]$. Then we get the co-induced module $M^G(\mathbb Q/\mathbb Z)$, which is acyclic by Shapiro's lemma.

2. The case of free modules follows by taking sums: $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z)$ turns sums into products, and a product of acyclic modules is acyclic. Indeed (by abstract nonsense): $(-)^G$ has a left adjoint $- \otimes_\mathbb Z \mathbb Z[G]$, hence preserves products (you can also easily prove this by hand). Since products of injectives are injective and products are exact, we can take them out of $H^i(G,-)$ as well.

3. The general case follows since any projective module is the summand of a free module.

Now consider the composition of the functors $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z) \colon \mathcal C^{\operatorname{op}} \to \mathcal C$ and $(-)^G \colon \mathcal C \to \operatorname{\underline{Ab}}$. The above argument shows that $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z)$ takes projectives to acyclics, hence there is a Grothendieck spectral sequence $$E_2^{pq} = H^p(G,\operatorname{Ext}^q_\mathbb Z(N,\mathbb Q/\mathbb Z)) \Rightarrow \operatorname{Ext}^{p+q}_{\mathbb Z[G]}(N,\mathbb Q/\mathbb Z).$$ But $\mathbb Q/\mathbb Z$ is injective as $\mathbb Z$-module, hence $\operatorname{Ext}^i_\mathbb Z(N,\mathbb Q/\mathbb Z) = 0$ for $i > 0$. Thus, the spectral sequence collapses on the $E_2$ page, and we conclude that $$H^i(G,\operatorname{Hom}_\mathbb Z(N,\mathbb Q/\mathbb Z)) = \operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z).$$ $\square$

Remark. If you don't like spectral sequences, you can also give a more concrete proof of the last part. Indeed, consider a free resolution as $\mathbb Z[G]$-modules $$\ldots \to P_1 \to P_0 \to N.$$ Since $\mathbb Z[G]$ is a free $\mathbb Z$-module, this resolution is also a free resolution as $\mathbb Z$-modules. Thus, it computes both $\operatorname{Ext}^i_\mathbb Z(N,\mathbb Q/\mathbb Z)$ and $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z)$. More specifically, the former can be computed as the cohomology of the complex $Q^\bullet$ given by $$0 \to \operatorname{Hom}_\mathbb Z(P_0,\mathbb Q/\mathbb Z) \to \operatorname{Hom}_\mathbb Z(P_1,\mathbb Q/\mathbb Z) \to \ldots,\tag{1}\label{Seq 1}$$ and then the latter can be computed from this complex as $$0 \to \operatorname{Hom}_{\mathbb Z[G]}(\mathbb Z,Q^0) \to \operatorname{Hom}_{\mathbb Z[G]}(\mathbb Z,Q^1) \to \ldots.\tag{2}\label{Seq 2}$$ Now (\ref{Seq 1}) is exact since $\mathbb Q/\mathbb Z$ is an injective $\mathbb Z$-module. By what we argued above, it is actually an acyclic resolution of $N^D = \operatorname{Hom}_\mathbb Z(N,\mathbb Q/\mathbb Z)$, so we can use it to compute group cohomology. Thus, $H^i(G,N^D)$ is computed as the cohomology of the complex $$0 \to \operatorname{Hom}_{\mathbb Z[G]}(\mathbb Z,Q^0) \to \operatorname{Hom}_{\mathbb Z[G]}(\mathbb Z,Q^1) \to \ldots,$$ which on the other hand computes $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z)$ by (\ref{Seq 2}). $\square$

Consider the short exact sequence $$0 \to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0.$$ Note that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q) = 0$ for all $i$: it is torsion since $N$ is torsion, but multiplication by $n \in \mathbb Z_{>0}$ is an isomorphism since it is so on $\mathbb Q$. Thus, the sequence above induces isomorphisms $$\operatorname{Ext}^{i-1}_\mathcal C(N,\mathbb Q/\mathbb Z) \stackrel\sim\to \operatorname{Ext}^i_\mathcal C(N,\mathbb Z).$$ Thus, we only have to show that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z) = H^i(G, N^D)$. I will treat the finite group case, and trust that the OP can carry out the limit argument to treat the profinite group $\hat {\mathbb Z}$ (assuming that you are working with discrete [topological] $G$-modules).

Note that if $P$ is a projective $\mathbb Z[G]$-module, then $\operatorname{Hom}_\mathbb Z(P,\mathbb Q/\mathbb Z)$ is an acyclic $\mathbb Z[G]$-module. Indeed, we prove this in three steps:

1. Let $P = \mathbb Z[G]$. Then we get the co-induced module $M^G(\mathbb Q/\mathbb Z)$, which is acyclic by Shapiro's lemma.

2. The case of free modules follows by taking sums: $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z)$ turns sums into products, and a product of acyclic modules is acyclic. Indeed (by abstract nonsense): $(-)^G$ has a left adjoint $- \otimes_\mathbb Z \mathbb Z[G]$, hence preserves products (you can also easily prove this by hand). Since products of injectives are injective and products are exact, we can take them out of $H^i(G,-)$ as well.

3. The general case follows since any projective module is the summand of a free module.

Now consider the composition of the functors $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z) \colon \mathcal C^{\operatorname{op}} \to \mathcal C$ and $(-)^G \colon \mathcal C \to \operatorname{\underline{Ab}}$. The above argument shows that $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z)$ takes projectives to acyclics, hence there is a Grothendieck spectral sequence $$E_2^{pq} = H^p(G,\operatorname{Ext}^q_\mathbb Z(N,\mathbb Q/\mathbb Z)) \Rightarrow \operatorname{Ext}^{p+q}_{\mathbb Z[G]}(N,\mathbb Q/\mathbb Z).$$ But $\mathbb Q/\mathbb Z$ is injective as $\mathbb Z$-module, hence $\operatorname{Ext}^i_\mathbb Z(N,\mathbb Q/\mathbb Z) = 0$ for $i > 0$. Thus, the spectral sequence collapses on the $E_2$ page, and we conclude that $$H^i(G,\operatorname{Hom}_\mathbb Z(N,\mathbb Q/\mathbb Z)) = \operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z).$$ $\square$

Remark. If you don't like spectral sequences, you can also give a more concrete proof of the last part. Indeed, consider a free resolution as $\mathbb Z[G]$-modules $$\ldots \to P_1 \to P_0 \to N.$$ Since $\mathbb Z[G]$ is a free $\mathbb Z$-module, this resolution is also a free resolution as $\mathbb Z$-modules. Thus, it computes both $\operatorname{Ext}^i_\mathbb Z(N,\mathbb Q/\mathbb Z)$ and $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z)$. More specifically, the former can be computed as the cohomology of the complex $$0 \to \operatorname{Hom}_\mathbb Z(P_0,\mathbb Q/\mathbb Z) \to \operatorname{Hom}_\mathbb Z(P_1,\mathbb Q/\mathbb Z) \to \ldots,\tag{1}\label{Seq 1}$$ and then the latter can be computed as the cohomology of $$0 \to \operatorname{Hom}_{\mathbb Z[G]}(P_0,\mathbb Q/\mathbb Z) \to \operatorname{Hom}_{\mathbb Z[G]}(P_1,\mathbb Q/\mathbb Z) \to \ldots.\tag{2}\label{Seq 2}$$ Now (\ref{Seq 1}) is exact since $\mathbb Q/\mathbb Z$ is an injective $\mathbb Z$-module. By what we argued above, it is actually an acyclic resolution $Q^\bullet$ of $N^D = \operatorname{Hom}_\mathbb Z(N,\mathbb Q/\mathbb Z)$, so we can use it to compute group cohomology. Thus, $H^i(G,N^D)$ is computed as the cohomology of the complex $$0 \to (Q^0)^G \to (Q^1)^G \to (Q^2)^G \to \ldots,$$ which on the other hand computes $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z)$ by (\ref{Seq 2}). $\square$