Here is an explanation Pavel Etingof has given to me in email. Thanks Pavel!

Every $\sigma\in S_{n}$ satisfies

$\sum\limits_{i}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\left[  x,y_{\sigma i}\right]  \ast\cdots\ast y_{\sigma n}$

$=n\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}$

$+\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\left[  x,y_{\sigma i}\right]  \ast\cdots\ast y_{\sigma n}$.

But since

$\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\underbrace{\left[
x,y_{\sigma i}\right]  }_{\substack{=x\ast y_{\sigma i}-y_{\sigma i}\ast
x\\\text{(since the inclusion of }\mathfrak{L}\\\text{into }\operatorname*{Sym}
\nolimits^{\ast}\mathfrak{L}\text{ is a morphism}\\\text{of Lie algebras)}
}}\ast\cdots\ast y_{\sigma n}$

$=\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\underbrace{\left(
x\ast y_{\sigma i}-y_{\sigma i}\ast x\right)  \ast\cdots\ast y_{\sigma n}
}_{\substack{=x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}-y_{\sigma i}\ast
x\ast\cdots\ast y_{\sigma n}\\\text{(by the induction hypothesis, since }i>1\text{)}}}$

$=\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma
i}\ast\cdots\ast y_{\sigma n}$

$-\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast
x\ast\cdots\ast y_{\sigma n}$

$= \sum\limits_{i>0}\left(  n-\left(  i+1\right)  +1\right)  \underbrace{y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma
\left(i+1\right)}\ast\cdots\ast y_{\sigma n}}_{=y_{\sigma1}\ast\cdots\ast
y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}}$

$-\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast
x\ast\cdots\ast y_{\sigma n}$ (here we substituted $i+1$ for $i$ in the first sum)

$=\sum\limits_{i>0}\left(  n-\left(  i+1\right)  +1\right)  y_{\sigma1}\ast\cdots\ast
y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$

$-\sum\limits_{i>1}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast
x\ast\cdots\ast y_{\sigma n}$

$=\left(  n-1\right)  y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits
_{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$,

this becomes

$\sum\limits_{i}\left(  n-i+1\right)  y_{\sigma1}\ast\cdots\ast\left[  x,y_{\sigma
i}\right]  \ast\cdots\ast y_{\sigma n}$

$=n\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast y_{\sigma
n}+\left(  n-1\right)  y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$

$-\sum\limits_{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast
y_{\sigma n}$

$=n\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast y_{\sigma
n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$

$-y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>1}y_{\sigma1}\ast
\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$

$=n\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast y_{\sigma
n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$

$-\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma
n}$.

Thus, (1.3.7.7) rewrites as

$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}=\left(
\text{symmetrized product of }x,y_{1},...,y_{n}\right)  $

$+\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}n\left[  x,y_{\sigma1}\right]
\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left(  n+1\right)
!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$

$-\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma
i}\ast x\ast\cdots\ast y_{\sigma n}$.

Since

$\left(  \text{symmetrized product of }x,y_{1},...,y_{n}\right)  $

$=\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}\left(  \sum\limits_{i>0}y_{\sigma1}
\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}+x\ast y_{\sigma
1}\ast\cdots\ast y_{\sigma n}\right)  $,

this simplifies to

$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}
=\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast
y_{\sigma n}$

$+\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}n\left[  x,y_{\sigma1}\right]
\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left(  n+1\right)
!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$.

Thus

$\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}n\left[  x,y_{\sigma1}\right]
\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left(  n+1\right)
!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$

$=\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}
-\dfrac{1}{\left(  n+1\right)  !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast
y_{\sigma n}$

$=\underbrace{\left(  \dfrac{1}{n!}-\dfrac{1}{\left(  n+1\right)  !}\right)
}_{=\dfrac{n}{\left(  n+1\right)  !}}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast
\cdots\ast y_{\sigma n}=\dfrac{n}{\left(  n+1\right)  !}\sum\limits_{\sigma}x\ast
y_{\sigma1}\ast\cdots\ast y_{\sigma n}$.

Divide this by $\dfrac{n}{\left(  n+1\right)  !}$ to obtain

$\sum\limits_{\sigma}\left[  x,y_{\sigma1}\right]  \ast y_{\sigma2}\ast\cdots\ast
y_{\sigma n}+\sum\limits_{\sigma}y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$

$=\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$.

In other words,

$0=\sum\limits_{\sigma}\left(  x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}
-y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\left[  x,y_{\sigma1}\right]
\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}\right)  $

$=\sum\limits_{\sigma}\left\lbrace  x,y_{\sigma1},...,y_{\sigma n}\right\rbrace  =\left(
n-1\right)  !\sum\limits_{i}\left\lbrace  x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n}
\right\rbrace  $

(here we used that $\left\lbrace  x_{1},...,x_{n+1}\right\rbrace  $ is symmetric in the
last $n-1$ variables, so that every $\sigma\in S_{n}$ satisfies $\left\lbrace
x,y_{\sigma1},...,y_{\sigma n}\right\rbrace  =\left\lbrace  x,y_{i},y_{1}
,...,\widehat{y_{i}},...,y_{n}\right\rbrace  $ for $i=\sigma1$).

Thus, $\sum\limits_{i}\left\lbrace  x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n}\right\rbrace
=0$, qed.