$\textbf{Lefschetz pencil.}$ As Lazzaro pointed out, the pencil you chose is not a Lefschetz pencil. A Lefschetz pencil is a general line in the dual space $(\mathbb P^3)^{\vee}$ intersecting the dual variety $\mathcal{D}$ of $F$ transversely at finitely many points (the number equals to $\deg \mathcal{D}$). To calculate this number, recall that the dual variety $\mathcal{D}$ is the image of the dual map
$$F\to (\mathbb P^3)^{\vee},\ x\mapsto (\frac{\partial F}{\partial X}(x),\frac{\partial F}{\partial Y}(x), \frac{\partial F}{\partial Z}(x),\frac{\partial F}{\partial W}(x)).$$
Take two general hyperplane planes $\{a_iL+b_iJ+c_iK+d_iH=0,i=1,2\}$ in the dual space $(\mathbb P^3)^{\vee}$. It follows that their intersections with $\mathcal{D}$ coincides with the solutions of
$$F=0,\ a_i\frac{\partial F}{\partial X}+b_i\frac{\partial F}{\partial Y}+c_i\frac{\partial F}{\partial Z}+d_i\frac{\partial F}{\partial W}=0,\ i=1,2$$
which contains $3\times 2\times 2=12$ points.
In particular, for a Lefschetz pencil of your Fermat cubic, there are exactly $12$ nodal cubic curves, so by applying your formula $\chi(F)=12-3=9.$
p.s., Usually we don't need to write down a Lefschetz pencil explicitly.
$\textbf{Special pencil.}$ However, it is still possible to calculate $\chi(F)$ via the pencil $\lambda X+Y=0$ that you chose. All you need is some topological properties of Eular characteristic $\chi$.
$\require{AMScd}$
\begin{CD}
\tilde{F} @>{\pi}>> \mathbb P^1\\
@V{\sigma}VV \\
F
\end{CD}
Consider the diagram where $\sigma$ blowup base locus (three points) so $\chi(\tilde{F})=\chi(F)+3$. $\pi$ is elliptic fibration with three singular fibers $C_{\omega^i},i=0,1,2$ are three cocurrent lines so
$\bullet$ $\chi(C_{\omega^i})=h^0+h^2=1+3=4$;
$\bullet$ The exclusion $U=\tilde{F}\setminus \{C_1,C_{\omega},C_{\omega^2}\}$ gives $\chi(\tilde{F})=\chi(U)+\sum \chi(C_{\omega^i})=\chi(U)+12$;
$\bullet$ The restriction $\pi_U:U\to V$ is a smooth fiber bundle, where $V=\mathbb P^1\setminus \{1,\omega,\omega^2\}$, so $\chi(U)=\chi(V)\chi(\text{elliptic curve})=0$.
Put these results together, you get $\chi(F)=9$.
Best Answer
Assume there is a single hyperplane section $X_0$ of $X$ which has worse singularity than just a node. Then any pencil containing $X_0$ is non-Lefschetz, and these pencils cover all the $\mathbb{P}^{n*}$.
Of course, if there is no such $X_0$ as above, this construction does not work. However, I think the only case when such $X_0$ does not exist is that of $X$ being a smooth quadric. In that case the projective dual variety to $X$ is yet another smooth quadric, and lines in $\mathbb{P}^{n*}$ tangent to it provide non-Lefschetz pencils that cover all the $\mathbb{P}^{n*}$.