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}}\frac1{n^2}}. 
\end{equation}
Clearly, $s_j^2\asymp2^j\frac1{(2^j)^2}\to0$ as $j\to\infty$. Also, for any integer $N>0$, 
\begin{equation}
\begin{aligned}
	S_N:=\sum_{N\le n<2N}\frac1{n^2}&=\sum_{N\le n<2N}\int_0^\infty du\,ue^{-nu} \\ 
	&=\int_0^\infty du\,u\sum_{N\le n<2N}e^{-nu} \\ 
	&=\int_0^\infty du\,u\frac{e^{-Nu}-e^{-2Nu}}{1-e^{-u}}	\\ 
	&=\int_0^\infty \frac{dz}z\,\frac{e^{-z}-e^{-2z}}{r(z/N)},  
\end{aligned}
\end{equation}
where $r(u):=\dfrac{1-e^{-u}}{u^2}$, which latter is decreasing in $u>0$. So, $S_N$ is decreasing in $N>0$ and hence $s_j$ is decreasing in $j=0,1,\dots$. 

So, by Theorem 1, p. 84 in [Kahane's book][1], 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 a random Fourier series of the form \eqref{1} is equivalent to the a.s. uniform convergence of the random series. $\quad\Box$

[1]: https://www.google.com/books/edition/Some_Random_Series_of_Functions/mquwmHIdT3UC?hl=en&gbpv=1&printsec=frontcover