$\newcommand\R{\mathbb R}
\renewcommand{\P}{\operatorname{\mathsf P}}\newcommand\E{\operatorname{\mathsf{E}}}\newcommand\Var{\operatorname{\mathsf{Var}}}$The best bound on $\Var \|X\|_2$ is about $n$.
Indeed,
\begin{equation*}
\Var \|X\|_2\le \E \|X\|_2^2=\sum_{i=1}^n \E X_i^2=n. \tag{1}\label{1}
\end{equation*}
On the other hand, suppose that
\begin{equation*}
\mu_X=q\,\mu_{aG}+p\,\mu_{bG},
\end{equation*}
where $\mu_Y$ denotes the distribution of a random vector $Y$, $G$ is a standard Gaussian random vector in $\R^n$, and
\begin{equation*}
a:=0,\quad b\in[1,\infty),\quad p:=\frac1{b^2},\quad q:=1-p.
\end{equation*}
Then the covariance matrix of $X$ is $pb^2I=I$.
Also, for all unit vectors $v\in\R^n$,
\begin{equation*}
\E (X\cdot v)^4=pb^4\E (G\cdot v)^4=3b^2=K^4
\end{equation*}
if $b=K^2/\sqrt3$ (assuming that $K^4\ge3$).
So, all your conditions hold.
However, $\E\|X\|_2^2=pb^2\E\|G\|_2^2=n$ and $\E\|X\|_2=pb\E\|G\|_2\le pb\sqrt{\E\|G\|_2^2}=pb\sqrt n=\sqrt n/b$ and hence
\begin{equation*}
\Var \|X\|_2=\E\|X\|_2^2-(\E\|X\|_2)^2\ge\Big(1-\frac1{b^2}\Big)n=\Big(1-\frac3{K^4}\Big)n; \tag{2}\label{2}
\end{equation*}
the inequality $\Var \|X\|_2\ge\big(1-\frac3{K^4}\big)n$ trivially holds if $K^4\not\ge3$.
Conclusion 1: The trivial upper bound $n$ on $\Var \|X\|_2$ in \eqref{1} is optimal, in the sense that it cannot be replaced by $cn$ for any real $c<1$, if $K$ is allowed to be large enough.
Nonetheless, the upper bound $n$ on $\Var \|X\|_2$ in \eqref{1} can be "second-order" improved as follows. Note that
\begin{equation*}
\E\|X\|_2^4=\E\Big(\sum_{i=1}^n X_i^2\Big)^2=\sum_{i,j=1}^n \E X_i^2 X_j^2
\le\sum_{i,j=1}^n \frac{\E X_i^4+\E X_j^4}2
=n\sum_{i=1}^n \E X_i^4\le n^2K^4.
\end{equation*}
So, by the log-convexity of $\E\|X\|_2^p$ in $p>0$,
\begin{equation*}
(\E\|X\|_2)^2\ge\frac{(\E\|X\|_2^2)^3}{\E\|X\|_2^4}\ge\frac{n^3}{n^2K^4}
=\frac n{K^4}.
\end{equation*}
So,
\begin{equation*}
\Var \|X\|_2=\E\|X\|_2^2-(\E\|X\|_2)^2\le\Big(1-\frac1{K^4}\Big)n, \tag{3}\label{3}
\end{equation*}
which is an improvement of \eqref{1}.
Conclusion 2: In view of \eqref{2} and \eqref{3}, the best upper bound on $\Var \|X\|_2$ is of the form
\begin{equation*}
\Big(1-\frac C{K^4}\Big)n
\end{equation*}
for some $C\in[1,3]$.
This refines Conclusion 1.
