This is one of those things that is immediate if you use regular conditional probabilities. For $\mathbb x \in \mathbb R^{\mathbb N}$, and a random variable $Y$, let $\mu^{\mathbb x}_Y$ denote the regular conditional probability distribution of $Y$ given $(X_1, X_2, \dots) = \mathbb x$. We also use $\mu_X$ to denote the joint law of the $X_i$ on $\mathbb R^n$.
By independence of $W_i$ from $X_i$, we have that the conditional distribution of $Z_n$ given $X_i = \mathbb x$ is just $\frac{1}{\sqrt n} \sum_{i = 1}^n W_i x_i$, where I use $x_i$ to denote the $i$’th component of $\mathbb x$.
Now we may use the following definition of weak conditional convergence: $Z_n \to Z$ weakly conditional on $X_i$ if for $\mu_X$-almost every $\mathbb x$, we have that $\mu^{\mathbb x}_{Z_n} \to \mu^{\mathbb x}_{Z}$ weakly in the usual sense.
It can be checked that this definition is equivalent to yours (feel free to ask for details!). Using this equivalent definition, it seems that your $Z_n$ converge to a normal random variable with random parameter parametrized by the $X_i$. But due to the LLN on the $X_i$, the parameter should be deterministic in the limit.