These are some thoughts concerning the Question: \begin{align*} &S_n(3)=\sum_{1\leq i\leq n}x_i^3< \sum_{i\geq 1}4^{-i}=1/3,\\ &S_n(1,2)=\sum_{1\leq i<j\leq n}x_i x_j^2< \sum_{1\leq i<j}4^{-(2j-i)}=1/45,\\ &S_n(2,1)=\sum_{1\leq i<j\leq n}x_i^2 x_j< \sum_{1\leq i<j}4^{-j}=1/9,\\ &S_n(1,1,1)=\sum_{1\leq i<j<k\leq n}x_i x_j x_k< \sum_{1\leq i<j<k}4^{-(k+j-i)}=1/135. \end{align*} Hence $$(x_1+x_2+\dots+x_n)^3=3(2S_n(1,1,1)+S_n(2,1)+S_n(1,2))+S_n(3)<\frac{7}{9}$$ which means that $C\leq \sqrt[3]{7/9}$.

On the other hand, note that $x_1=1/4$ and $x_i=1/4^{i-1}$ for $2\leq i\leq n$ satisfy the required conditions and $$x_1+x_2+\dots+x_n=\frac{7}{12}-\frac{1}{3\cdot 4^{n-1}}.$$ Therefore $C\geq 7/12$.

These are some thoughts concerning the Question. We have that \begin{align*} &S_n(3)=\sum_{1\leq i\leq n}x_i^3< \sum_{i\geq 1}4^{-i}=1/3,\\ &S_n(1,2)=\sum_{1\leq i<j\leq n}x_i x_j^2< \sum_{1\leq i<j}4^{-(2j-i)}=1/45,\\ &S_n(2,1)=\sum_{1\leq i<j\leq n}x_i^2 x_j< \sum_{1\leq i<j}4^{-j}=1/9,\\ &S_n(1,1,1)=\sum_{1\leq i<j<k\leq n}x_i x_j x_k< \sum_{1\leq i<j<k}4^{-(k+j-i)}=1/135. \end{align*} Hence $$(x_1+x_2+\dots+x_n)^3=3(2S_n(1,1,1)+S_n(2,1)+S_n(1,2))+S_n(3)<\frac{7}{9}$$ which means that $C\leq \sqrt[3]{7/9}$.

On the other hand, note that $x_1=1/4$ and $x_i=1/4^{i-1}$ for $2\leq i\leq n$ satisfy the required conditions and $$x_1+x_2+\dots+x_n=\frac{7}{12}-\frac{1}{3\cdot 4^{n-1}}.$$ Therefore $C\geq 7/12$.