Consider a sequence $Z_n$ of random vectors in $R^d$ such that $Z_n\to N(0_d,I_d)$ in distribution.
Consider a sequence of random matrices $T_n\in R^{d\times d}$, not independent of $Z_n$, such that $T_nT_n^\top\to C$ for a deterministic positive semi-definite $C\in R^{d\times d}$.
I am interested in whether $T_n Z_n$ converges in distribution to $N(0_d, C)$.
Remark 1: This is Slutsky's theorem if $d=1$.
Remark 2: If $T_n\to^d A$ and $C=AA^T$, then we may apply the continuous mapping theorem to the function $f(T,z)=Tz$. But here we only assume convergence of $T_nT_n^\top$ to $C$.
$T_n Z_n$ converges in distribution to $N(0_d, C)$ not true. A counterexample is as follows: $T_n$ is a rotation in $O(d)$ for all $n$ such that the first row is $Z_n/\|Z_n\|$, so that $T_nT_n^\top = I_d$ but $T_n Z_n$ has zero in coordinates $2,3,...,d$.
So let us add another assumption to avoid this rotation situation:
Question: Is it always true that $T_n Z_n$ converges in distribution to $N(0_d, C)$ provided that $T_n$ is lower triangular with $1$ on the diagonal?
The answer to your last question is positive. Indeed, if $C$ is positive definite (which is your implicit assumption here), then the Cholesky factor $A$ in $C = AA^T$ depends on $C$ continuously, so $T_n\to A$ in distribution (and actually in probability, since $A$ is non-random).