Function $W \mapsto f(W)$ is not convex when $n \ge 2$.

Take for example when $n=2$ and fix $x=1$.

If function $f(W_1,W_2)$ is convex then it must also be convex when restricted to a linear subspace.

Restricting to $W_1 = \begin{pmatrix} x & 0 \\ 0 & 0 \end{pmatrix}$ and $W_2 = \begin{pmatrix} y & 0 \\ 0 & 0 \end{pmatrix}$ then gives $f(W_1,W_2) = x^2 y^2$, which is not convex in (x,y) (this can be checked by computing its Hessian matrix).

Function $W \mapsto f(W)$ is not convex when $n \ge 2$.

Take for example when $n=2$ and fix $x=1$.

If function $f(W_1,W_2)$ is convex then it must also be convex when restricted to a linear subspace.

Restricting to $W_1 = \begin{pmatrix} u & 0 \\ 0 & 0 \end{pmatrix}$ and $W_2 = \begin{pmatrix} v & 0 \\ 0 & 0 \end{pmatrix}$ then gives $f(W_1,W_2) = u^2 v^2$, which is not convex in $(u,v)$ (this can be checked by computing its Hessian matrix).

Function $W \mapsto f(W)$ is not convex.

Already the simplest case, take $n=1$ (matrices $W_i$ and vector $x$ become scalars) and fix $x=1$: then function $f$ becomes the product of the squares of its arguments, which is not convex unless there is only one factor in the product (for example $(x,y) \mapsto x^2 y^2$ is not convex, as can be checked by computing its Hessian matrix).

Function $W \mapsto f(W)$ is not convex when $n \ge 2$.

Take for example when $n=2$ and fix $x=1$.

If function $f(W_1,W_2)$ is convex then it must also be convex when restricted to a linear subspace.

Restricting to $W_1 = \begin{pmatrix} x & 0 \\ 0 & 0 \end{pmatrix}$ and $W_2 = \begin{pmatrix} y & 0 \\ 0 & 0 \end{pmatrix}$ then gives $f(W_1,W_2) = x^2 y^2$, which is not convex in (x,y) (this can be checked by computing its Hessian matrix).