6. Given a regular tetrahedron $P-ABC$ with all edge lengths equal to $1$, $L$, $M$, and $N$ are the midpoints of edges $PA$, $PB$, and $PC$ respectively. Then the area of the section formed by the plane $LMN$ cutting the circumsphere of the regular tetrahedron is . $\qquad$
6. $\frac{\pi}{3}$.

From the conditions, we know that plane $L M N / /$ plane $A B C$, and the ratio of the distances from point $P$ to plane $L M N$ and plane $A B C$ is $1: 2$.

Let $H$ be the center of the face $A B C$ of the regular tetrahedron $P-A B C$, and let $P H$ intersect plane $L M N$ at point $K$.
Then $P H \perp$ plane $A B C, P K \perp$ plane $L M N$.
Thus, $P K=\frac{1}{2} P H$.
The regular tetrahedron $P-A B C$ can be considered as a regular tetrahedron, and let $O$ be its center (i.e., the center of the circumscribed sphere), then $O$ lies on $P H$, and by the properties of a regular tetrahedron, we know that $O H=\frac{1}{4} P H$.

Combining $P K=\frac{1}{2} P H$, we know that $O K=O H$, which means point $O$ is equidistant from plane $L M N$ and plane $A B C$.

This indicates that the cross-sectional circles formed by the circumscribed sphere of the regular tetrahedron when intersected by plane $L M N$ and plane $A B C$ are of equal size.

Therefore, the area of the required cross-section is equal to the area of the circumcircle of $\triangle A B C$, which is $\pi\left(\frac{A B}{\sqrt{3}}\right)^{2}=\frac{\pi}{3}$.