Here is an idea for the proof of the statement. Feedback appreciated.
The class $\mathscr{H}:=\{x\mapsto e^{i u\cdot x},u \in \mathbb{Q}^d\}$ is a convergence determining class on $\mathbb{R}^d$ (reference, p. 347). This means that $\mu_n \to \mu$ weakly if $\int_{\mathbb{R}^d} h d\mu_n\to \int_{\mathbb{R}^d} h d\mu$ for all $h \in \mathscr{H}$ for $n \to \infty$. In our case $\mu_n=P_{Z_n},\,\mu=P_Z$ where $Z_n$ is our standardized process and $Z\sim \mathcal{N}_d(0,Q)$. So by IID-ness
$$\int_{\mathbb{R}^d}e^{i u\cdot x}dP_{Z_n}=E[e^{iu\cdot Z_n}]=(E[e^{iu \cdot\frac{(X_1-\mu)}{\sqrt{n}}}])^n$$
We now use the argument by Taylor series $e^{iy}=1+iy-y^2/2+...$ as in the proof for the case $d=1$. We get (recall $E[X_1-\mu]=0$)
$$E[e^{iu \cdot\frac{(X_1-\mu)}{\sqrt{n}}}]=1-\frac{1}{2n}E[(u\cdot (X_1-\mu))^2]+...$$
we now use the fact that for $u,v \in \mathbb{R}^d$ we have $(u\cdot v)^2=u^Tvv^Tu$ where $^T$ is the vector transpose. Recall $E[(X_1-\mu)(X_1-\mu)^T]=Q$. So
$$(E[e^{iu \cdot\frac{(X_1-\mu)}{\sqrt{n}}}])^n=\bigg(1-\frac{1}{2n}u^TQu+...\bigg)^n$$
The term $u^TQu$ is a scalar. By sending $n \to \infty$ we obtain $\lim_{n \to \infty}\int_{\mathbb{R}^d}e^{i u\cdot x}dP_{Z_n}=e^{-\frac{1}{2}u^TQu}$ for all $u \in \mathbb{R}^d$. We can see that the rhs is the characteristic function of the law $\mathcal{N}_d(0,Q)$. Thus
$$\lim_{n \to \infty}\int_{\mathbb{R}^d}e^{i u\cdot x}dP_{Z_n}=\int_{\mathbb{R}^d}e^{i u\cdot x}dP_{Z},\quad \forall u \in {\mathbb{R}^d}$$
This estalishes $Z_n\to Z$ in distribution.
