n-cocycles of finite abelian groups from cohomology group
3 years later.
My recent paper (https://arxiv.org/abs/1703.03266) answers this question.
More precisely, let $\mathbb{k}$ be an algebraically closed field of characteristic zero. By $\mathbb{k}^*$ we denote the multiplicative group $\mathbb{k}-\{0\}$. Let $G=\mathbb{Z}_{m_{1}}\times\cdots \times\mathbb{Z}_{m_{n}}$ where $m_i|m_{i+1}$ for $1\le i\le n-1$ and $(B_{\bullet},\partial_{\bullet})$ be its normalized bar resolution. Applying $\text{Hom}_{\mathbb{Z}G}(-,\mathbb{k}^*)$ one gets a complex $(B^*_{\bullet},\partial^*_{\bullet})$.
Let $\alpha:=(\alpha_{11},\dots,\alpha_{1n},\dots,\alpha_{k1},\dots,\alpha_{kn})$ where $0\leq\alpha_{ij}<m_{j}$ for $1\leq i\leq k$. For $1\leq r\leq n$, $[a,b]\subseteq[1,k]$, $a,b\in\mathbb{N}$ and $\alpha$, denote
$$ \eta_{r,[a,b]}^{\alpha}:=\left\{ \begin{array}{ll}[\frac{\alpha_{br}+\alpha_{b-1,r}}{m_{r}}]\cdots[\frac{\alpha_{a+1,r}+\alpha_{ar}}{m_{r}}]&\;\;\;a-b\text{ odd}\\ [\frac{\alpha_{br}+\alpha_{b-1,r}}{m_{r}}]\cdots[\frac{\alpha_{a+2,r}+\alpha_{a+1,r}}{m_{r}}]\alpha_{ar}&\;\;\;a-b\text{ even} \end{array}\right. $$ The following $\omega\in \text{Hom}_{\mathbb{Z}G}(B_k,\mathbb{k}^*)$ \begin{eqnarray} &&\omega([g_1^{\alpha_{11}}\cdots g_n^{\alpha_{1n}},\dots,g_1^{\alpha_{k1}}\cdots g_n^{\alpha_{kn}}])\\\notag &=&\prod_{l=1}^{k}\prod_{\begin{array}{ccc}1\leq r_{1}<\cdots<r_{l}\leq n\\\lambda_1+\cdots+\lambda_l=k,\lambda_1\text{ odd}\\\lambda_i\ge1\text{ for }1\le i\le l\end{array}}\zeta_{m_{r_1}}^{(-1)^{\sum_{1\leq i<j\leq l}\lambda_{i}\lambda_{j}}\eta_{r_{1},[a_{1},b_{1}]}^{\alpha}\cdots\eta_{r_{l},[a_{l},b_{l}]}^{\alpha}a_{r_1^{\lambda_1}\cdots r_l^{\lambda_l}}} \end{eqnarray} where $a_{l}=1,b_{l}=\lambda_{l},\dots,a_{1}=\lambda_{2}+\cdots+\lambda_{l}+1,b_{1}=\lambda_{1}+\cdots+\lambda_{l}=k$ and $0\leq a_{r_1^{\lambda_1}\cdots r_l^{\lambda_l}}<m_{r_1}$ for $1\leq r_1<\cdots<r_l\leq n$ makes a complete set of representatives of $k$-cocycles of the complex $(B_{\bullet}^*,\partial_{\bullet}^*)$. Moreover, $$\text{H}^k(G,\mathbb{k}^*)=\prod_{r=1}^n\mathbb{Z}_{m_r}^{\sum_{j=1}^k(-1)^{k+j}\binom{n-r+j-1}{j-1}}.$$
An algorithmic way to describe the standard n-cocycle (cocycles respect to the bar resolution) for abelian groups is given in Lyndon's paper The cohomology theory of group extensions (It is no more than the LHS spectral sequences in a very particular case).As Mariano said, first you need to describe the n-cocycles for cyclic groups (constructing a chain homomorphism between the cyclic resolution and the Bar resolution). After that you only need to follow the ideas of Lyndon's paper in order to construct explicitly the standard n-cocycles.
By the way, in Lyndon's paper also there is an explicit description of $H^{n}(A,U(1))\cong H^{n+1}(A,\mathbb{Z})$ for $A$ a finite abelian group.
(Not a complete answer.) Let me say what I have tried in $n=3$ (this is only a partial answer I already know).
Given a generic finite abelian group $G=\mathbb{Z}_{N^{(1)}} \times \cdots \times \mathbb{Z}_{N^{(k)}}$. I compute that 3rd-cohomology group is $$ H^3(G,R/\mathbb{Z}) \simeq \! \bigoplus_{1 \leq i < j < l \leq k} \! Z_{N^{(i)}} \oplus Z_{{\gcd} (N^{(i)},N^{(j)})} \oplus Z_{{\gcd} (N^{(i)},N^{(j)},N^{(l)})} \, . $$ ($R/\mathbb{Z}=U(1)$)
I also compute the corresponding 3-cocycles are: $$ \omega_{{I}}^{(i)}(A,B,C) = \exp \left( \frac{2 \pi i p^{(i)}_{{I}}}{N^{(i)\;2}} \; a^{(i)}(b^{(i)} +c^{(i)} -[b^{(i)}+c^{(i)}]) \right) \\ \omega_{{II}}^{(ij)}(A,B,C) = \exp \left( \frac{2 \pi i p_{{II}}^{(ij)}}{N^{(i)}N^{(j)}} \; a^{(i)}(b^{(j)} +c^{(j)} - [b^{(j)}+c^{(j)}]) \right) \\ \omega_{{III}}^{(ijl)} (A,B,C) = \exp \left( \frac{2 \pi i p_{{III}}^{(ijl)}}{{\gcd}(N^{(i)}, N^{(j)},N^{(l)})} \; a^{(i)}b^{(j)}c^{(l)} \right) $$ with $p^{(i)}_{{I}}$ labels the group element $Z_{N^{(i)}}$, $p_{{II}}^{(ij)}$ labels the group element in $Z_{{\gcd} (N^{(i)},N^{(j)})}$ and $p_{{III}}^{(ijl)}$ labels the group element in $Z_{{\gcd} (N^{(i)},N^{(j)},N^{(l)})}$.
I am still interested in knowing other $n$, especially $n=2,4,5$. Please reply in the specific forms in $n=3$ as I did for $H^3(G,R/\mathbb{Z})$ and $\omega$. Thank you.