Let $P\in \mathrm{GL}_n(\mathbb{Q})$ be positive definite, and $(\det P)^{\frac{1}{n}}\in \mathbb{Q}$.

Question. Can one always write $P=(\det P)^{\frac{1}{n}} AA^*$, with $A\in \mathrm{SL}_n(\mathbb{Q}[i])$ ?

Note that $P$ cannot always be written as $P=AA^*$, with $A\in \mathrm{GL}_n(\mathbb{Q}[i])$. Indeed, this would imply $\det P=a\overline{a}$, where $a=\det A$, but $\det P$ is not always a sum of two rational squares.

As well, note that $P$ can always be diagonalised, as $P=LDL^\top$, with $L,D\in \mathrm{GL}_n(\mathbb{Q})$, and $L$ lower trianglular with 1s on the main diagonal, $D$ diagonal. Thus $P$ can be assumed to be diagonal.

Probably, we cannot always take $A\in \mathrm{GL}_n(\mathbb{Q})$, as the diagonal entries of $P$ are sums of $n$ squared norms of entries of $A$, and this would not always work for $n=2$ if all the entries of $A$ are in $\mathbb{Q}$.

I don't even know the answer for $n=2$. On the other hand, one can e.g. have

$$ P=\begin{pmatrix} 2 & 0 \\ 0 & 8 \end{pmatrix} = 4AA^*, \quad A=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\\-1&1\end{pmatrix}. $$

A less trivial example, which also shows that the conditions on $A$ and $P$ may be weakened:

$$ P=\begin{pmatrix} 7 & 0 \\ 0 & 7/2 \end{pmatrix} = AA^*, \quad A=\begin{pmatrix} 2+i&1+i\\ 1&-\frac{3+i}{2} \end{pmatrix}. $$ Here $(\det P)^\frac{1}{2}=\frac{7}{\sqrt{2}}\not\in\mathbb{Q}$.

Thus, in general, let's consider the decomposition $\det P=\zeta_P\sigma_P$, where $\sigma_P$ the "biggest" part of $\det P$ representable as a norm in $\mathbb{Q}[i]$ (in the latter example $\sigma_P=1/2$, $\zeta_P=7^2$). Thus the modified question is as follows.

Question′. $P\in \mathrm{GL}_n(\mathbb{Q})$ be positive definite, and satisfying $\zeta_P^{\frac{1}{n}}\in \mathbb{Q}$. Can one always write $P=AA^*$, with $A\in \mathrm{GL}_n(\mathbb{Q}[i])$ ?