The answer is that $n_L=81$. The proof was originally due to Fano and the literature can be found in page 40 from Tyurin's *Five lectures on three-dimensional varieties, 1972*. For convenience of others and for my own benefit, I will rewrite the proof below. The notation is the same as above.

**I. Degree of the Scroll**: Let $I$ be the incidence variety of the pair $(L,x)$ such that $x\in L$
$\require{AMScd}$
\begin{CD}
I @>{\varphi}>> X\\
@V{p}VV \\
F
\end{CD}

with $p:I\to F$ is a $\mathbb P^1$-bundle over the curve $F$, and $\varphi(I)$ is the scroll of the surface swept out by lines in quartic threefold $X$, which is singular at points where lines are incident to each other. According to a standard Schubert calculus computation in Tennison's *On the quartic threefold, 1974*, the degree of surface $\varphi(I)$ is $320$, or equivalently, $$\varphi(I)=80H\tag{1}\label{1}$$ in $A^1(X)$, where $H$ is a hyperplane section.

**II. $n_L$ as Intersection Number**: Let $\Gamma, L\subset I$ be a section and a fiber of $p:I\to F$, respectively, then among the points of the intersection $$\varphi(L)\cdot \varphi(\Gamma)\tag{2}\label{2}$$ is one point of the intersection of $L$ with $\Gamma$ and **the remaining points are intersection of $L$ with other lines in the family**. Therefore it suffices to find a special section such that the intersection number $(\ref{2})$ is easy to compute.

Let $H'$ be a general hyperplane containing a line $L$ in $X$, then $S=H'\cap X$ is a smooth surface with $L$ as the unique line. So in the Chow ring $A^*(X)$, $$\varphi(I)\cdot S=L+C\tag{3}\label{3}$$ with $C$ the residue curve which does not contain a line. Therefore $\varphi^{-1}(\varphi(I)\cdot S)=L+\varphi^{-1}(C)$, and

*Proposition: $\varphi^{-1}(C)$ is a section of the bundle $p:I\to F$. It follows that
$$n_L=L\cdot C-1.$$*

**III. Final Computation**: Intersect with $L$ on both side of $(\ref{3})$, and use $(\ref{1})$, we have

$$80(L+C')\cdot L=80(L^2+3)=L^2+L\cdot C$$

where $C'$ is a plane cubic intersecting $L$ transversely at $3$ points. Now use the fact that $S$ has trivial canonical bundle, so $L$ has self-intersection $(-2)$ in $S$, which leads to $n_L=L\cdot C-1=81$.

distinctfrom $L$ is 81. $\endgroup$3more comments