Yes, and the good news are that there isn't anything to compute. By a result of Kempf a line bundle $L$ on $\mathbb{P}^3$ descends to the quotient if and only if the stabilizer $S_x$ of each point $x\in \mathbb{P}^3$ acts trivially on the fiber $L_x$. Now $S_4$ acts on $\mathcal{O}(1)$, and $S_x$ act on $\mathcal{O}(1)_x$ through a character $\chi _x:S_x\rightarrow \mathbb{C}^*$, and on $\mathcal{O}(1)^{\otimes k}$ through the character $\chi _x^{k}$. The abelian subquotients of $S_4$ have order $2,3$ or $4$, hence taking $k=12$ kills all the characters, and therefore $\mathcal{O}(1)^{\otimes k}$ descends.

Yes, and the good news are that there isn't anything to compute. By a result of Kempf a line bundle $L$ on $\mathbb{P}^3$ descends to the quotient if and only if the stabilizer $S_x$ of each point $x\in \mathbb{P}^3$ acts trivially on the fiber $L_x$. Now $S_4$ acts on $\mathcal{O}(1)$, and $S_x$ act on $\mathcal{O}(1)_x$ through a character $\chi _x:S_x\rightarrow \mathbb{C}^*$, and on $\mathcal{O}(1)^{\otimes k}$ through the character $\chi _x^{k}$. The abelian subquotients of $S_4$ have order $2,3$ or $4$, hence taking $k=12$ kills all the characters, and therefore $\mathcal{O}(1)^{\otimes 12}$ descends.

Yes, and the good news are that there isn't anything to compute. By a result of Kempf a line bundle $L$ on $\mathbb{P}^3$ descends to the quotient if and only if the stabilizer $S_x$ of each point $x\in \mathbb{P}^3$ acts trivially on the fiber $L_x$. Now $S_4$ acts on $\mathcal{O}(1)$, and $S_x$ act on $\mathcal{O}(1)_x$ through a character $\chi _x:S_x\rightarrow \mathbb{C}^*$, and on $\mathcal{O}(1)^{\otimes k}$ through the character $\chi _x^{k}$. The abelian subquotients of $S_4$ have order $2,3$ or $4$, hence taking $k=12$ kills all the characters, and therefore $\mathcal{O}(1)$ descends.

Yes, and the good news are that there isn't anything to compute. By a result of Kempf a line bundle $L$ on $\mathbb{P}^3$ descends to the quotient if and only if the stabilizer $S_x$ of each point $x\in \mathbb{P}^3$ acts trivially on the fiber $L_x$. Now $S_4$ acts on $\mathcal{O}(1)$, and $S_x$ act on $\mathcal{O}(1)_x$ through a character $\chi _x:S_x\rightarrow \mathbb{C}^*$, and on $\mathcal{O}(1)^{\otimes k}$ through the character $\chi _x^{k}$. The abelian subquotients of $S_4$ have order $2,3$ or $4$, hence taking $k=12$ kills all the characters, and therefore $\mathcal{O}(1)^{\otimes k}$ descends.