Let \begin{equation*} F(x):=\sum_{n=1}^{\infty}\frac{X_n}n\,\cos nx. \tag{1}\label{1} \end{equation*} For $j=0,1,\dots$, let \begin{equation*} s_j:=\sqrt{\sum_{2^j\le n<2^{j+1}}E\Big(\frac{X_n}n\Big)^2}=\sqrt{\sum_{2^j\le n<2^{j+1}}\frac1{n^2}}. \end{equation*}
For any integer $N>0$ and
\begin{equation*}
S_N:=\sum_{N\le n<2N}\frac1{n^2}
\end{equation*}
we have $S_{N+1}-S_N=-\dfrac1{N^2}+\dfrac1{(2N)^2}+\dfrac1{(2N+1)^2}<0$.
So, $S_N$ is decreasing in $N=1,2,\dots$ and hence $s_j$ is decreasing in $j=0,1,\dots$.
Also, $s_j^2\asymp2^j\frac1{(2^j)^2}\to0$ as $j\to\infty$.
So, by Theorem 1, p. 84 in Kahane's book, the function $F$ is continuous almost surely (a.s.). (The condition in that theorem that $s_j$ be decreasing seems possible to relax.) As noted on p. 48 of Kahane's book, the a.s. continuity of the sum of a random Fourier series of the form \eqref{1} is equivalent to the a.s. uniform convergence of the random series. $\quad\Box$