This is a partial answer. Consider the case $k=1$. The parametric solutions are given by $(x,y,z)=\left(2ab(a^2+b^2),a^4-b^4,2ab(a^2-b^2)\right)$, where $\gcd(a,b)=1$, $a>b>0$ and $a+b$ is odd. In the triples $(x,y,z)$ generated by this parametrization, it is not necessarily always $x\leq y$. For example, $(20,15,12)$ is a triple generated by this parametrization, and $(15,20,12)$ will not be generated. This parametrization generates only the unique solutions. So we can have two possible cases, either $0<a^2-b^2\leq 2ab$ or $0<2ab\leq a^2-b^2$.
The sum of the triples = $x+y+z=4a^3b+a^4-b^4 \leq n$.
Let us consider the first case. Denote the number of solutions by $A_1^1(n)$ and $A_1^1=\lim_{n \to \infty}\frac{A_1^1(n)}{\sqrt{n}}$.
$$0<a^2-b^2\leq 2ab \implies \sqrt{2}-1\leq\frac{b}{a}=x<1$$
Number of pairs $(a,b)$ satisfying above constraints and also $4a^3b+a^4-b^4 \leq n$ can be approximated by
\begin{align}
\begin{split}
p(n)&\approx\int_{\frac{b}{a}=\sqrt{2}-1}^{1}\int_{4a^3b+a^4-b^4 \leq n}a \ da\ db\\
&=\int_{x=\sqrt{2}-1}^{1}\int_{0<a^4\leq \frac{n}{1+4x-x^4}}a \ da\ dx \\
&=\frac{\sqrt{n}}{2}\int_{x=\sqrt{2}-1}^{1}\frac{1}{\sqrt{1+4x-x^4}}\ dx
\end{split}
\end{align}
The natural density of coprime numbers is $\frac{6}{\pi^2}$. The condition $gcd(a,b)=1$ will contribute a multiplication factor of $\frac{6}{\pi^2}$. Among all the coprime pairs $(a,b)$ such that $a>b>0$, fraction of which have odd value of $a+b$ is $\frac{2}{3}$. We have to multiply this factor.
\begin{align}
\begin{split}
A_1^1 &=\lim_{n \to \infty}\frac{A_1^1(n)}{\sqrt{n}}\\
&= \lim_{n \to \infty} \frac{2}{3} \cdot \frac{6}{\pi^2} \cdot\frac{p(n)}{\sqrt{n}}\\
&= \frac{2}{\pi^2} \int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{1+4x-x^4}} \\
\end{split}
\end{align}
Similarly, it can be shown that for the other case exactly the same limit exists. So, $$A_1=2A_1^1=\frac{4}{\pi^2} \int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{1+4x-x^4}} \approx 0.1277999513464289...$$
Using a similar approach it can be shown that
\begin{align}
\begin{split}
A_2&=\frac{2}{\pi^2} \left[ \int_{x=0}^{\sqrt{2}-1}\frac{dx}{\sqrt{3-6x^2-x^4}} +\int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{4x+6x^2+4x^3-1-x^4}} \right]\\
& \approx 0.1036994744684913...
\end{split}
\end{align}
\begin{align}
\begin{split}
A_{13}&=\frac{52}{7\pi^2} \int_{x=(\sqrt{13}-3)/2}^{(\sqrt{26}-1)/5}\frac{dx}{\sqrt{20x+36x^2-5-7x^4}}\approx 0.1038855856479065...
\end{split}
\end{align}
Best Answer
$$(a^2+b^2+c^2+d^2)^2=3(a^4+b^4+c^4+d^4)$$ Let assume $c=a+b,$ then we get $$a^2+ab+b^2=d^2$$ A parametric solution is given $$(a,b,c,d)=(p^2-q^2, 2pq+q^2, p^2+2pq, q^2+pq+p^2)$$ where $p,q$ are any integer.