We induct on $n$. Notice that $n = 1$ is trivial and in fact is an identity. Now suppose $n$ works. We have $x_1x_2\ldots x_n = \frac{1}{x_{n + 1}} = t^{-1}$, where $t = x_{n + 1}$, and thus\[\sum_{i = 1}^nx_i\sqrt{x_1^2 + \cdots + x_i^2}\ge\frac{(n + 1)\sqrt{n}}{2}t^{-\frac{2}{n}},\]while $x_1^2 + \cdots + x_n^2 + x_{n + 1}^2 \ge n + 1$ so that\[x_{n + 1}\sqrt{x_1^2 + \cdots + x_n^2 + x_{n + 1}^2}\ge t\sqrt{n + 1}.\]Finally,\[\sum_{i = 1}^{n + 1}x_i\sqrt{x_1^2 + \cdots + x_i^2}\ge\frac{(n + 1)\sqrt{n}}{2}t^{-\frac{2}{n}} + t\sqrt{n + 1} > \frac{n\sqrt{n + 1}}{2}t^{-\frac{2}{n}} + t\sqrt{n + 1} = \sqrt{n + 1}\left(\frac{n}{2}t^{-\frac{2}{n}} + t\right).\]Combine with\[\frac{n}{2}t^{-\frac{2}{n}} + t\ge\frac{n}{2} + 1,\]from AM-GM, to complete the inductive step. In fact, this is a strict inequality for $n > 1$. Thus the best constant, $a_n$, satisfies\[\frac{(n + 1)\sqrt{n}}{2} < a_n \le \sum_{i = 1}^n \sqrt{i}\]for $n > 1$. I conjecture that $a_n\approx \frac{2}{3}n^{\frac{3}{2}}$, i.e., the equality case is closer to $(1, 1, \ldots, 1)$ than to what we proved above. We can actually prove $a_n > cn^{\frac{3}{2}}$, where $c = \frac{1}{2}e^{\frac{1}{4}}\approx 0.64201$. We proceed in exactly the same way; for $n = 1$, we have $a_n > c$, which is clear. Then, for the inductive step we have\[\sum_{i = 1}^{n + 1} x_i\sqrt{x_1^2 + \cdots + x_i^2}\ge a_nt^{-\frac{2}{n}} + t\sqrt{n + 1} = \sqrt{n + 1}\left(\frac{n}{2}\left(\frac{2a_n}{n\sqrt{n + 1}}t^{-\frac{2}{n}}\right) + t\right) \ge \sqrt{n + 1}\left(\frac{n}{2} + 1\right)\left(\frac{2a_n}{n\sqrt{n + 1}}\right)^{\frac{\frac{n}{2}}{\frac{n}{2} + 1}},\]by similar AM-GM, and the RHS is at least\[\sqrt{n + 1}\left(\frac{n}{2} + 1\right)\left(\frac{2cn\sqrt{n}}{n\sqrt{n + 1}}\right)^{\frac{n}{n + 2}} \ge c(n + 1)\sqrt{n + 1},\]where the last is equivalent to\[\left(2c\sqrt{\frac{n}{n + 1}}\right)^{\frac{n}{n + 2}}\ge 2c\left(\frac{n + 1}{n + 2}\right)\iff (2c)^n\left(1 - \frac{1}{n + 1}\right)^{\frac{n}{2}}\ge (2c)^{n + 2}\left(1 - \frac{1}{n + 2}\right)^{n + 2}\]or\[\frac{\left(1 - \frac{1}{n + 1}\right)^{\frac{n}{2}}}{\left(1 - \frac{1}{n + 2}\right)^{n + 2}}\ge (2c)^2.\]And of course\[\left(1 - \frac{1}{n + 1}\right)^n > e^{-1} > \left(1 - \frac{1}{n + 2}\right)^{n + 2},\]so it's clear since $(2c)^2 = e^{\frac{1}{2}}$. Thus, our asymptotics are from $\frac{e^{\frac{1}{4}}}{2}n^{\frac{3}{2}}$ to $\frac{2}{3}n^{\frac{3}{2}}$. We might be able to do better if we write\[x_{n + 1}\sqrt{x_1^2 + \cdots + x_n^2 + x_{n + 1}^2} \ge t\sqrt{t^2 + nt^{-\frac{2}{n}}},\]which is a little stronger.

\begin{align*} \sum_{i = 1}^{n}x_i\sqrt{x^2_1+x^2_2+\cdots x^2_i} &= \sum_{i=1}^n i\cdot\left(\frac{x_i}{i}\cdot\sqrt{x_1^2+x_2^2+...+x_i^2}\right) \\ &\ge \frac{n(n+1)}{2} \left(\prod_{i=1}^n \frac{x_i^i}{i}(x_1^2+x_2^2+...+x_i^2)^{0.5i} \right) ^{\frac{2}{n(n+1)}} \\ &\ge \frac{n(n+1)}{2} \left(\prod_{i=1}^n \frac{x_i^i}{i^i} \cdot i^{0.5i} (x_1x_2...x_i) \right) ^{\frac{2}{n(n+1)}} \\ &= \frac{n(n+1)}{2} \left( \frac{\prod\limits_{i=1}^n x_1x_2...x_{i-1}\cdot x_i^{i+1} }{\prod\limits_{i=1}^n i^{0.5i}} \right) ^{\frac{2}{n(n+1)}} \\ &\ge \frac{n(n+1)}{2} \left( \frac{(x_1x_2...x_n)^{n+1}}{\prod\limits_{i=1}^n n^{0.5i}} \right) ^{\frac{2}{n(n+1)}} \\ &= \frac{n(n+1)}{2} \left( \frac{ 1 }{ n^{0.25n(n+1)} } \right) ^{\frac{2}{n(n+1)}} \\ & = \frac{n+1}{2}\cdot\sqrt{n} \end{align*}where we have used weight AM-GM and AM-GM in the second and third inequalities.

\begin{align*} \sum_{i=1}^nx_i\sqrt{x_1^2+\cdots+x_i^2} &\geqslant \sum_{i=1}^nx_i\cdot \frac{x_1+\cdots+x_i}{\sqrt{i}} \\ & = \sum_{i=1}^n\sum_{j=1}^i\frac{x_ix_j}{\sqrt{i}} \\ & \geqslant \frac{1}{\sqrt{n}}\sum_{i=1}^n\sum_{j=1}^i x_ix_j \\ & = \frac{1}{2\sqrt{n}}\sum_{1\leqslant j\leqslant i\leqslant n}2x_ix_j \\ & = \frac{1}{2\sqrt{n}}\left(\sum_{i=1}^nx_i^2 + \left(\sum_{i=1}^nx_i\right)^2\right) \\ & \geqslant \frac{1}{2\sqrt{n}}\left[n\cdot \sqrt[n]{x_1^2\cdots x_n^2} + \left(n\cdot \sqrt[n]{x_1\cdots x_n}\right)^2\right] \\ & = \frac{n+1}{2}\sqrt{n}.\quad\square \end{align*}