We can show that River Li’s bound $c_n\ge \frac{16}{27}(n-1)n^2=b_n$ is tight as follows.
Given any real $x_1,\dots x_n$ with $\sum x_i=1$ for each $i$ put $y_i=x_i-\tfrac 1n$. Then $x_i-x_j=y_i-y_j$ for each $i$ and $j$, $\sum y_i=0$, and
$$\sum x_i^2=\sum \left(y_i+\frac 1n\right)^2=\sum y_i^2+\frac 2n y_i+\frac 1{n^2}=$$ $$
\sum y_i^2+ \frac 2n\left(\sum y_i\right)+n\frac 1{n^2}=\sum y_i^2+\frac 1{n}.$$
Then
$$\sum_{1\le i<j\le n}(x_{i}-x_{j})^2=n\sum_{i=1}^{n}x^2_{i}-1=n\sum y_i^2=nY\ge 0.$$
Thus we have to show that
$$b_n(Y+\tfrac 1n)^3\ge n\left(\sum_{i=1}^{n}|y_{i+1}-y_{i}|\right)^2Y.$$
For this we need the following auxiliary result.
For each natural $k$ and each point $z=(z_1,\dots, z_k)\in \Bbb R^k$ let $\|z\|=\sqrt{\sum_{i=1}^k z_i^2}$, $B_k=\{z\in\Bbb R^k: \|z\|\le 1\}$ be the unit ball in the space $\Bbb R^k$, and $f:B_k\to\Bbb R$ be a function such that $$f(z)=\sum_{i=1}^{k}|z_{i+1}-z_{i}|$$
for each $z\in B_k$, where $z_{k+1}=z_1$. Since the set $B_k$ is compact and the function $f_k$ is continuous, it attains its maximum $M_k$ at some point $t=(t_1,\dots,t_k)\in B_k$. Clearly, $\sum_{i=1}^{n}|y_{i+1}-y_{i}|\le \sqrt{Y}M_k$.
By the inequality between arithmetic and quadratic means, we have
$$M_k=\sum_{i=1}^{k}|t_{i+1}-t_{i}|\le \sum_{i=1}^{k}|t_{i+1}|+|t_{i}|=$$ $$2\sum_{i=1}^{k}|t_i|\le
2\sqrt{k}\sqrt{\sum_{i=1}^{k}t_i^2}=2\sqrt{k}\|t\|\le 2\sqrt{k}.$$
It is easy to check that if $k$ is even then the equality above is attained when $t_i=\tfrac {(-1)^i}{\sqrt{k}}$ for each $i$.
Now assume that $k>1$ is odd. Suppose to the contrary that all $t_i$ are non-zero. We claim that there exists a natural $i$ such that both $t_i$ and $t_{i+1}$ (in the circular order) have the same sign. Indeed, otherwise the sign of each $t_i$ is $(-1)^{i-1}\operatorname{sign} t_1$. Since $k$ is odd, $t_1$ and $t_k$ have the same sign, a contradiction. Let $t_j,\dots, t_l$ be a longest circular subsequence of consecutive $t_i$’s with the same sign. Let $t_m$ be a number in this subsequence with the smallest absolute value. Let a point $t’\in\Bbb R^k$ has the same coordinates as $t$, but only its $m$-th coordinate is zero, and $$\lambda=\frac {\|t\|}{\|t’\|}>1.$$ Then $\|\lambda t’\|=\|t\|\le 1$, so $\lambda t’\in B_k$. On the other hand,
it is easy to check that $f(t’)\ge f(t)$, so $f(\lambda t’)=\lambda f(t’)>f(t)$, a contradiction with the maximality of $f(t)$. Thus there exists $m$ such that $t_m=0$. Then similarly to the above case of even $k$ we can show that
$$f(t)\le 2\sum_{i=1}^{k}|t_i|\le 2\sqrt{k-1}\sqrt{\sum_{i=1}^{k}t_i^2}=2\sqrt{k-1}\|t\|\le 2\sqrt{k-1}.$$
It is easy to check that if the equality above is attained when $t_i=\tfrac {(-1)^i}{\sqrt{k-1}}$ for each $i<k$ and $t_k=0$. Thus $M_k=2\sqrt{k-1}$.
Since $n$ is odd, $$\left(\sum_{i=1}^{n}|y_{i+1}-y_{i}|\right)^2\le YM_n^2=4Y(n-1).$$
So it remains to show that
$b_n(Y+\tfrac 1n)^3\ge 4(n-1)nY^2$
$\frac{16}{27}(n-1)n^2(Y+\frac 1n)^3\ge 4(n-1)nY^2$
$4n(Y+\frac 1n)^3\ge 27Y^2$
$4(nY+1)^3\ge 27n^2Y^2$
Put $y=nY\ge 0$. We have to show that
$4(y+1)^3\ge 27y^2$
$4y^3-15y^2+12y+4\ge 0$
$(y-2)^2(4y+1)\ge 0$.
