For the special case of $n=2$, I think I have a complete answer to Question': We get a factoring $P=AA^*$ with $A\in GL_2(\mathbb{Q}[i])$ iff $\det(P)$ is the sum of two rational squares.
For the proof, I will only focus on diagonal matrices. Let $a,b> 0$ be the diagonal entries of a diagonal matrix $P$. The condition that $AA^*$ is diagonal is equivalent to the rows (or columns) of $A$ being orthogonal (in the complex scalar product), which means that $A$ has the form
$$A=\begin{pmatrix}u & v\\ z\bar{v} & -z\bar{u}\end{pmatrix}$$
and $a=u\bar{u}+v\bar{v}$, $b=z\bar{z}(u\bar{u}+v\bar{v})=z\bar{z}a$. We can always find $u$ and $v$ to fulfill the first condition, since every nonnegative rational number can be written as a sum of four squares of rational numbers by Lagrange's four squares theorem (and the equation is actually just $a=\Re(u)^2+\Im(u)^2+\Re(v)^2+\Im(v)^2$)
The second condition can be fulfilled iff $\frac{b}{a}=z\bar{z}=\Re(z)^2+\Im(z)^2$, i.e. iff $\frac{b}{a}$ is the sum of two squares of rational numbers. Multiplying both sides by $a^2$ implies that this is equivalent to $\det(P)=ab=(\Re(z)a)^2+(\Im(z)a)^2$.
(If $\det(P)=0$, i.e. $P$ semidefinite, this also holds since we can then just take one of the rows to be $0$ and the other as in the four squares theorem)
Using this method we can find some factorizations in higher dimensions, however I do not have a characterization yet. For example: If we can split the numbers along the diagonal into pairs, such that the quotient of each pair is a sum of two squares, we can apply it to each pair individually.
Thinking about it, there may be a way to generalize.
Claim: A diagonal matrix $M\in GL_n(\mathbb{Q})$ with rational entries $d_1\geq d_2\geq \dots \geq d_n>0$ for some $k\leq n$ can be factored as $M=AA^*$ with $A\in GL_n(\mathbb{Q}[i]$ if and only if $\det(M)$ is the sum of two squares (in particular, if $M\in SL_n(\mathbb{Q})$).
Proof: The only if part is clear by the determinant multiplication formula and noting that $z\bar{z}=\Re(z)^2+\Im(z)^2$ for $z\in \mathbb{Q}[i]$.
For the other direction, I will use, but not prove, the following fact:
Fact (1): If $n>0$ is a rational number, then the quadratic form $q:\mathbb{Q}^4\to\mathbb{Q}_0^+$, $q(s,t,u,v)=s^2+t^2+n(u^2+v^2)$ is surjective. In particular, every nonnegative rational number $x$ can be written as $x=r^2+s^2+n(t^2+u^2)$ for some rational numbers $r,s,t,u$.
Using this as a black box, we iteratively construct an orthogonal basis of $\mathbb{Q}[i]$ with $\langle v_i\mid v_i\rangle=d_i$ for $i=1,\dots , n$ in the following way:
Step 0: Using the four squares theorem (or $n=1$ in Fact), we find some numbers $r_0,s_0,t_0,u_0$ such that $r_0^2+s_0^2+t_0^2+u_0^2=d_1$. Set $v_1=(r_0+is_0)e_1+(t_0+iu_0)e_2$. Further, set $w_{1}=e_1$. Now, $\langle v_1\mid v_1\rangle=d_1$
Iteration for $k=1,\dots , n-2$: Let $v_1,\dots , v_k$ be already constructed. If $e_{k+1}\in \mathrm{span}(v_1,\dots ,v_k)$, set $b_{k}:=w_{k}$ and otherwise set $b_{k}:=e_{k+1}$. Now we already have $b_k\perp v_i$ for $i=1\dots , k-1$.
Set $$w_{k+1}:=b_k-\sum_{j=1}^k\frac{\langle v_j\mid b_k\rangle}{\langle v_j\mid v_j\rangle}v_j=b_k-\frac{\langle v_k\mid b_k\rangle}{\langle v_k\mid v_k\rangle}v_k$$
(In other words: Extend $v_1,\dots , v_k$ by $w_{k+1}$to an orthogonal basis of $\mathrm{span}(e_1,\dots , e_{k+1})$).
$w_{k+1}$ is now some non-zero vector, i.e. $\langle w_{k+1}\mid w_{k+1}\rangle=N_{k+1}$ for some positive rational number $N_{k+1}$. We can therefore (by Fact) find some rational numbers $r_{k+1},s_{k+1},t_{k+1},u_{k+1}$ such that $$N(r_{k+1}^2+s_{k+1}^2)+t_{k+1}^2+u_{k+1}^2=d_{k+1}$$ holds.
Set $$v_{k+1}:=(r_{k+1}+is_{k+1})w_{k+1}+(t_{k+1}+iu_{k+1})e_{k+2}$$
Then $v_{k+1}$ is orthogonal to the vectors $v_{1},\dots , v_{k}$, because $w_{k+1}$ and $e_{k+2}$ are, and we get $$\langle v_{k+1}\mid v_{k+1}\rangle=(r_{k+1}^2+s_{k+1}^2)\langle w_{k+1}\mid w_{k+1}\rangle+(t_{k+1}^2+u_{k+1}^2)\langle e_{k+2}\mid e_{k+2}\rangle=N(r_{k+1}^2+s_{k+1}^2)+(t_{k+1}^2+u_{k+1}^2)=d_{k+1}$$
from $w_{k+1}\perp e_{k+2}$.
Last step (k=n-1): Now, there is only a one dimensional vector space remaining for the last vector. In this vector space, there is already a vector $w_{n}$ with $\langle w_n\mid w_n\rangle=d_1\cdot d_2\cdots \cdot d_{n-1}$ and coefficients in $\mathbb{Q}[i]$ (the external product of the other $n-1$ vectors). This vector can now only scaled by a complex number to the correct length, i.e. $d_{n}/(d_{n-1}\cdots d_{1})$ must be the norm of a complex number, i.e. a sum of squares. But this condition is equivalent to $\det(P)=d_1\cdot d_2\cdots d_{n-1}\cdot d_n$ being a sum of squares (multiply by a square). Since we did not need any further conditions, this concludes the proof.
This means that $P$ can be split as $P=AA^*$ in $GL_n({Q}[i])$ iff the "trivial" restriction that the determinant needs to split is fulfilled.
This is in particular the case if $P\in SL_n(\mathbb{Q})$. If $\det(P)=q^n\cdot (x^2+y^2)$ for some rational numbers $q,x,y$, we can bring $q$ (but not necessarily all of $\det$) to the front and find our decomposition that way; and for $\det(P)=q^n$, we can choose $A$ in $SL_n(\mathbb{Q})[i]$.
As a remark: If $d_k=0$ for some $k$ (and all $j>k$), we can by the same method also find a decomposition, but now every row and column starting with $k$ will be $0$, for the other rows, we use the same construction as before.
Since it was asked in a comment, I am adding the idea of a proof for Fact (1) here:
The idea is to apply the local-global principle (Hasse-Minkowski) to the quadratic form $v^2+w^2+n(x^2+y^2)-cz^2=0$. It is quite obvious that this has a nontrivial rational root, iff $c$ is in the image of the $q$ mentioned in fact (1)($z$ can not be $0$ then, so we can divide by $z^2$). It is also easy to see that $z=1,v=\sqrt{c}$ is a root in $\mathbb{R}$.
The p-adic fields are a bit more involved, and to make things easier, we may assume (by simple quadratic substitution in x and y or z) that $n$ and $c$ are squarefree integers (if $c=0$, the problem would be trivial; and we assumed $n>0$).
$p=2$ is the messiest part: We can't use Hensel's Lemma here, however it is well-known that every integer congruent to $1 \pmod 8$ has a square root in $\mathbb{Z}_2$. Hence we set $z=1$ and check for each of the $6$ possible values for $c \pmod 8$ ($0$ and $4$ are not squarefree) and each of the $6$ possible values for $n \pmod 8$ that we can get numbers for $w,x,y$ such that $w^2+n(x^2+y^2)-q=7\pmod 8$ and take as $v$ a square root of $-(w^2+n(x^2+y^2)-q)$.
For $p>2$, we can basically use the proof of Lagrange's theorem, however we need some preparation.
There are 3 cases: $p\nmid q$, $p\mid q$ and $p\mid n$, and $p\mid q$ and $p\nmid n$.
In the first case, we set $x=y=0$, and only need to solve $v^2+w^2-qz^2=0$
In the second case, we set $v$ and $w$ to $0$ and then divide by $p$ to get an equation of the form $nx^2+ny^2-cz^2=0$ where $p$ is not a divisor of the coefficients.
In the third case, we set $y=z=0$ to get $v^2+w^2+nx^2=0$
In any case, it suffices to show that for $p\nmid a,b,c$ the equation $ax^2+by^2+cz^2=0$ has some nontrivial solution in $\mathbb{Q}_p$. So, we set $z=1$ (which guarantees that a solution will be nontrivial). By the same pigeonhole-principle argument as in Lagrange's 4 squares theorem, we get some $x',y'$ such that $a{x'}^2+b{y'}^2+c=0\pmod p$ and since $p\nmid c$, $x'\neq 0\pmod p$ or $y'\neq q$, so we can use Hensel's Lemma to lift this to a solution of $ax^2+by^2+c=0$ in $\mathbb{Q}_p$.
Now, we have a nontrivial root in each local field, so by the local-global principle the quadratic form also has a non trivial root in $\mathbb{Q}$.
