This is **not an answer**, but only an attempt to retrieve, and present, the early result mentioned by the OP when $n = 2$. Hopefully, this will trigger further input from MO readers.

The following is immediate.

**Claim 1.** The image of $[0, 1]^n$ by any matrix in $GL_n(\mathbb{Z})$ is a fundamental domain of the torus $\mathbb{R}^n/\mathbb{Z}^n$.

The converse is **false**, as shown for instance by the parallelotope $ [0, 1](1, \frac{1}{2}) + [0, 1](0,1) \subset \mathbb{R}^2$.

But this is somehow, the only kind of exceptions we can get when $n = 2$.

**Claim 2.** A parallelotope in $\mathbb{R}^2$ is a fundamental domain of the torus $\mathbb{R}^2/\mathbb{Z}^2$, if and only if, it is the image of $[0, 1](1, x) + [0, 1](0, 1)$ by some matrix in $GL_n(\mathbb{Z})$ for some $x \in [0, 1]$. Equivalently, if and only if the matrix of the parallelotope can be reduced to $\begin{pmatrix} 1 & x \\
0 & 1 \end{pmatrix}$ by means of elementary integral row operations for some $x \in [0, 1]$.

By integral elementary row operation I mean, adding or subtracting one row to the other, or negating a row.

**Proof of Claim 2.**
If a parallelotope $P$ has the form mentioned in Claim 2, then it is easy to check that $P$ is a fundamental domain. We suppose now that $P$ is a fundamental domain.
Let us write $P = [0, 1]v_1 + [0, 1]v_2$ where $v_1$ and $v_2$ are the rows of $V = \begin{pmatrix} v_{11} & v_{12} \\ v_{21} & v_{22}\end{pmatrix}$.
By hypothesis, we have $P + \mathbb{Z^n} = \mathbb{R}^n$ and $\dot{P} \cap (P + z) = \emptyset$ for every non-zero $z$ in $\mathbb{Z}^n$. From this, we infer that $\vert \det(V) \vert = \mu(P) = 1$ where $\mu$ denotes the Lebesgue measure on $\mathbb{R}^2$. Without loss of generality, we can replace $V$ by $AV$ for any $A \in GL_2(\mathbb{Z})$. In particular, we can try to reduce $V$ by means of elementary row operations. If we don't manage to cancel one of the coefficients of the first column of $V$, then these coefficients can be made arbitrarily small, say less than $\frac{1}{2}$, and positive. This is impossible because $P$ would then be contained in a vertical strip of width smaller than $1$ and hence could not surject onto $\mathbb{R}^2/\mathbb{Z}^2$ via the natural map $\mathbb{R}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}^2$. Therefore we can assume, without loss of generality, that $v_{21} = 0$, $v_{11} > 0$, $v_{22} > 0$ and $0 \le v_{12} < v_{22}$.
Necessarily, $v_{11} \ge 1$, since otherwise $P$ would be contained in a vertical strip of width smaller than $1$. If $v_{11} > 1$ and $v_{12} > 0$, then $v_1 - (1, 0)$ belongs to the interior of $P$, a contradiction. We are left with two cases.

Case 1. $v_{11} = 1$. As $\det(V) = 1$, this implies $v_{22} = 1$.

Case 2. $v_{12} = 0$. Then $P$ is a rectangle, and since $P$ surjects onto $\mathbb{R}^2/\mathbb{Z}^2$, we necessarily have $v_{11} = v_{22} = 1$.

In each case, the matrix $V$ has the desired form.

**Edit.** Elaborating on the above result, we can try to get a more general statement with

**Claim $n$.** A parallelotope in $\mathbb{R}^n$ is a fundamental domain of the torus $\mathbb{R}^n/\mathbb{Z}^n$ if and only if the matrix $V = V(P)$ whose rows are the generators of $P$ can be reduced by means of integral elementary row operations, including negating rows, to a matrix of the form $$\begin{pmatrix}
1 & a_{11} & \cdots & \cdots & a_{nn} \\ 0 & 1 & a_{23} & \cdots & a_{2n} \\ 0 & 0 & 1 &
\ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & 1 \end{pmatrix}$$ with $a_{ij} \in [0, 1]$ whenever $j > i$.

This may be exactly what the OP already mentioned in his question. This is certainly what Yoav Kallus refers to in his answer. (So I invite the OP to elaborate on his expectations regarding the description of fundamental domains, e.g., do we seek for uniqueness of the matrix? In which ways this description could be insufficient?).

**Proof Proposal for Claim $n$.** We can reduce $V$ to an upper triangular form via induction, arguing as in the case $n = 2$. Using induction on $n$ and the fact that we can suppose $\det(V) = 1$, we obtain the result by projecting $P$ on a direct factor $\mathbb{R}^{n -1}$ of $\mathbb{R}^n$.

As pointed out by Yoav Kallus, Claim $n$ is equivalent to Minkowski's conjecture on lattice cube-tiling of $\mathbb{R}^n$, so the reasoning sketched above is certainly flawed for $n > 3$. But, as Yoav Kallus already said, the conjecture holds true by Hajós' theorem. I found the lecture notes of P. Shor very helpful to understand the link between OP's question and Minkowski's cube-tiling conjecture. We learn in Section 3 that Minkowski established the result for $n = 3$ but failed to generalize it. Section 4 is dedicated to Hajós' theorem.