I have cross-posted this question on Math Overflow here.
For a finite group $G$, the rational group ring $\mathbb{Q}[G]$ has a Wedderburn decomposition: $$ \mathbb{Q}[G] \cong \prod_{i=1}^k M_{n_i}(D_i), $$ where $D_i$ is a division algebra whose center is a number field, and $M_{n_i}(D_i)$ is the ring of $n_i \times n_i$ matrices over $D_i$. I will refer to the factors of this product as the simple factors of $\mathbb{Q}[G]$. Each simple factor corresponds to an irreducible $\mathbb{Q}$-representation of $G$.
My question is: given $n$ and a primitive $d$th root of unity $\zeta_d$, for which finite groups $G$ does $M_n(\mathbb{Q}(\zeta_d))$ appear as a simple factor of $\mathbb{Q}[G]$ corresponding to a faithful representation of $G$?
It is well known that if $G$ is cyclic of order $d$, then the unique faithful irreducible representation of $G$ has $\mathbb{Q}(\zeta_d)$ as its corresponding simple factor.
Moreover, if $G$ is a Heisenberg group over $\mathbb{Z}/p\mathbb{Z}$, then it has a unique faithful irreducible representation; computer experiments (using the GAP package wedderga) seem to suggest that the corresponding simple factor is $M_p(\mathbb{Q}(\zeta_p))$.
What other examples are there?
I will prove the statement when $G$ is the Heisenberg group over $\mathbb F_p$ (or more generally, over $\mathbb F_q$ for some prime power $q$). First of all, your claim about uniqueness of the faithful representation is false, see: Character table of modular Heisenberg groups. I will outline a construction of them.
Fix some embedding $\chi\colon \mathbb F_q\hookrightarrow\mathbb C^\times$. Consider the normal subgroup $$N=\begin{pmatrix}1&\mathbb F_q&\mathbb F_q\\ &1\\&&1\end{pmatrix}\subset G=\begin{pmatrix}1&\mathbb F_q&\mathbb F_q\\ &1&\mathbb F_q\\&&1\end{pmatrix},$$ which has a character $\theta\colon N\to\mathbb C^\times$. $$\theta\begin{pmatrix}1&a&b\\ &1\\&&1\end{pmatrix}:=\chi(b).$$ Now let $\pi:=Ind_N^G\theta$. An explicit Mackey computation shows $\pi$ is indeed irreducible. Moreover, we may explicitly compute the character of $\theta$, which is supported on $N$, as: $$ch_\pi\begin{pmatrix}1&a&b\\ &1\\&&1\end{pmatrix}=\sum_{x\in\mathbb F_q}\chi(ax+b),$$ so in particular when $a=0$ we have $ch_\pi(a)=q\chi(x)$. In particular, since the character is defined over $\mathbb Q(\zeta_q)$, the division algebra $D_\pi$ corresponding to $\pi$ must be over $\mathbb Q(\zeta_q)$.
Moreover, from the construction of $\pi$ it is clear that it can be defined over $\mathbb Q(\zeta_q)$, so $D=\mathbb Q(\zeta_q)$.