Let me myself comment briefly the calculation of $H^d[U(1)^n,U(1)]$ from another angle, the Chern-Simons theory.
It is known that the relation between Chern-Simons action $CS(A)$:
\begin{equation}
e^{2\pi i CS(A)}=\exp\Bigl[ 2\pi i k \int_{X_3} A \wedge dA\Bigr]
\end{equation}
as a partition function of $U(1)$ Chern-Simons action $CS(A)$ on a 3-manifold $X_3$.
For a compact gauge group $G$ the Chern-Simons actions are classified by
an element [Ref:D-W]
\begin{equation}
k\in H^4(BG,\mathbb{Z})
\end{equation}
The CS actions $CS(A)$ for
a compact gauge group $G$ are in one-to-one correspondence
with the elements of the cohomology
group $H^4 (B{ G}, \mathbb{Z})$ of the classifying space
$B{ G}$ with integer coefficients $\mathbb{Z})$.
For $G=U(1)$, CS actions $CS(A)$ is classified by
\begin{equation}
H^4(B(U(1)),\mathbb{Z}) \cong \mathbb{Z}
\end{equation}
, and the element $\mathbb{Z}$ is simply the integer $k$ (level $k$)
appearing in the expression for the $U(1)$ Chern-Simons action on the 3-manifold $X_3$.
For $K_{n\times n}$ Chern-Simons action $CS(A)$ with $U(1)^n$ gauge group:
\begin{equation}
e^{2\pi i CS(A)}=\exp\Bigl[ 2\pi i K_{ij} \int_{X_3} A_{i} \wedge dA_{j}\Bigr]
\end{equation}
For $G=U(1)^n$, CS actions $CS(A)$ is classified by
\begin{equation}
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
\end{equation}
---
Let's shortly sketch the proof of $H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$
\begin{equation}
H^d(BU(1),\mathbb{Z}) \simeq \begin{cases}
\mathbb{Z} & \text{if $d\in$ even} \newline
0 & \text{if $d\in$ odd}
\end{cases}
\end{equation}
For $H^d(B(U(1)^n),\mathbb{Z})$ with $n>1$, we can use the Kunneth
formula, because the classifying space of the product group
$U(1)^n$ is the same as the product of the the factorization of classifying spaces.
That is, $B(U(1)^n) = B(U(1)^{n-1}) \times BU(1)$.
The calculation is analogue to $H^3[\mathbb{Z}_p^n,U(1)]$ case.
It is easier to show that for discrete abelian group:
\begin{equation}
\begin{cases}
H^3[\mathbb{Z}_p,U(1)]=\mathbb{Z}_p, \text{(confirmed)}\newline
H^3[\mathbb{Z}_p^2,U(1)]=(\mathbb{Z}_p)^3 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^3,U(1)]=(\mathbb{Z_p})^7 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^n,U(1)]=\mathbb{Z_p}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} \text{(confirmed)} \newline
\end{cases}
\end{equation}
Since the group $\mathbb{Z}$ is torsion free, however, the terms
due to torsion products $Tor$ vanish in this case (thanks to the discussion: https://mathoverflow.net/questions/128203/torsion-product-torr-1-closed).
To summarize, the contribution of $H^4(B(U(1)^n),\mathbb{Z})$, comes from
$n$ terms of the form $H^4(BU(1))\simeq \mathbb{Z}$. These
label the different CS actions of diagonal $K_{n\times n}$ matrix type:
\begin{equation}
\end{equation}
\begin{equation}
{\sum_{i=1}^n {K_{(ii)}} \epsilon^{\kappa\sigma\rho} {A_\kappa^{(i)}} \partial_{\sigma} A^{(i)}_{\rho} }
\end{equation}
In addition, there are $\frac{1}{2} n(n-1)$ terms
of the form $H^2(BU(1)) \simeq \mathbb{Z}$ which label
the CS actions of off-diagonal $K_{n\times n}$ matrix type:
\begin{equation}
{\sum_{i\neq j}^n {K_{(ij)}} \epsilon^{\kappa\sigma\rho} {A_\kappa^{(i)}} \partial_{\sigma} A^{(j)}_{\rho} }
\end{equation}
So overall, $H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$ (q.e.d.)
---
Importantly, this classification includes the case of finite gauge groups $H$.
**The isomorphism for a finite gauge groups $H$**.
\begin{equation}
H^d(B{H},{\mathbb{Z}}) \simeq H^d({H},{\mathbb{Z}}) ,
\end{equation}
Moreover, the universal coefficient theorem(UCThm) shows the isomorphism
\begin{equation}
H^{d} (H, \mathbb{Z}) \simeq H^{d-1} ({ H}, U(1))
\qquad \forall n>1.
\end{equation}
We have then:
\begin{equation}
H^4 ({BH}, \mathbb{Z}) \simeq H^3 ({ H}, U(1))
\end{equation}
-------
It is this relation: $\begin{equation}
H^4 ({BH}, \mathbb{Z}) \simeq H^3 ({ H}, U(1))
\end{equation}$
allure me from this finite gauge groups $H$ result,
\begin{equation}
\begin{cases}
H^3[\mathbb{Z}_p,U(1)]=\mathbb{Z}_p, \text{(confirmed)}\newline
H^3[\mathbb{Z}_p^2,U(1)]=(\mathbb{Z}_p)^3 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^3,U(1)]=(\mathbb{Z_p})^7 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^n,U(1)]=\mathbb{Z_p}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} \text{(confirmed)} \newline
\end{cases}
\end{equation}
to prompt the guessed result of:
\begin{equation}
\begin{cases}
H^3[U(1),U(1)]=\mathbb{Z}, \newline
H^3[U(1)^2,U(1)]=(\mathbb{Z})^3 \text{(to be checked)} \newline
H^3[U(1)^3,U(1)]=(\mathbb{Z})^7 \text{(to be checked)} \newline
H^3[U(1)^n,U(1)]=\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} \text{(to be checked)} \newline
\end{cases}
\end{equation}
However, the result
$
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$
already tells me the right way to view this $U(1)^n$ gauge symmetry C-S theory should be classified by $H^4(B(U(1)^n),\mathbb{Z})$
as
\begin{equation}
\begin{cases}
H^4(B(U(1)),\mathbb{Z})=\mathbb{Z}, \newline
H^4(B(U(1)^2),\mathbb{Z}) \cong \mathbb{Z}^{3} \newline
H^4(B(U(1)^3),\mathbb{Z}) \cong \mathbb{Z}^{6} \newline
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)} \newline
\end{cases}
\end{equation}
This is what I should ask in **(Q1)**.
So, instead, what I should really ask is, apart from **(Q1)(Q2)**:
**(Q3)**
whether there is a **symmetry breaking picture, such that one can obtain the result of
$H^3[\mathbb{Z}_p^n,U(1)]$ of a discrete $\mathbb{Z}_p^n$ group from a large continuous group, say, from $U(1)^n$ broken down to $\mathbb{Z}_p^n$?
So that, for example, this guessed $H^3[U(1)^n,U(1)]$ broken down to a subgroup picture works**
\begin{equation}
H^3[U(1)^n,U(1)] (\text{guessed})\to H^3[\mathbb{Z}_p^n,U(1)]
\end{equation}
as
\begin{equation}
\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} (\text{guessed})\to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}
\end{equation}
We know that however
\begin{equation}
H^4(B(U(1)^n),\mathbb{Z}) \to H^3[\mathbb{Z}_p^n,U(1)]
\end{equation}
broken down from
\begin{equation}
\mathbb{Z}^{n+\frac{1}{2}n(n-1)} \to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}
\end{equation}
This does not work, where we cannot simply replacing $\mathbb{Z}$ to $\mathbb{Z}_p$ by symmetry breaking from $U(1)$ to $\mathbb{Z}_p$. It has been known that one may need a symmetry breaking of C-S action with a nonAbelian continuous group broken down to subgroup $Z_p^n$ to produce the all known elements of $H^3[\mathbb{Z}_p^n,U(1)]$.
So that goes back to my guessed proposal:
\begin{equation}
\begin{cases}
H^3[U(1),U(1)]=\mathbb{Z}, \newline
H^3[U(1)^2,U(1)]=(\mathbb{Z})^3 \text{(to be checked)} \newline
H^3[U(1)^3,U(1)]=(\mathbb{Z})^7 \text{(to be checked)} \newline
H^3[U(1)^n,U(1)]=\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} \text{(to be checked)} \newline
\end{cases}
\end{equation}
PS. I have to apologize what I had mentioned may be intriguing, the question turns out to overlap different math fields. Instead of directly answering the questions **(Q1)(Q2)**, I now address the questions differently.
--------------
Ref:
[Ref:D-W]:Robbert Dijkgraaf, Edward Witten, Topological Gauge Theories and Group Cohomology, Commun. Math. Phys. 129 (1990), 393