A cancellation property for permutations?

Let $n$ be some integer greater than 2. Since the number of even and odd permutations in $S_n$ is the same we have $\sum_{\sigma\in S_{n}}(-1)^{\ell(\sigma)}=0$ therefore the contribution of $\sum_{\sigma\in S_{n}}(-1)^{\ell(\sigma)}\left(\sum_{i=1}^n i^2\right)$ is zero. It remains to show that $$\sum_{\sigma\in S_{n}}(-1)^{\ell(\sigma)}\sum_{i=1}^n i\sigma(i)=0.$$ Notice that if we write $P(x)=\det\left(x^{ij}\right)_{i,j=1}^n$ then this sum is simply $P'(1)$. However the order of vanishing of $P$ at $1$ is $\binom{n}{2}$ (notice that the matrix is pretty much a Vandermonde matrix) and this is greater than $1$ since $n>2$, therefore $P'(1)=0$.

Too long to fit in comments.

Remark(?).
$$\sum_{\sigma\in S_n}(-1)^{\ell(\sigma)}\sum_{i=1}^ni^{(any~real~number)}\sigma(i)=0.$$

That probably follows from Darij's comment. Here again: n>2 ( n=2 is really an exception).

The Python code below checks it. (One can use https://colab.research.google.com/ for free - you even do not need to install anything on your comp - use browser and code runs on google's servers: )

import numpy as np
import time

def inversion(permList): # http://code.activestate.com/recipes/579051-get-the-inversion-number-of-a-permutation/
    """
    Description - This function returns the number of inversions in a
                  permutation.
    Preconditions - The parameter permList is a list of unique positve numbers.

    Postconditions - The number of inversions in permList has been returned.

    Input - permList : list
    Output - numInversions : int
    """
    if len(permList)==1:
        return 0
    else:
        numInversion=len(permList)-permList.index(max(permList))-1
        permList.remove(max(permList))
        return numInversion+inversion(permList)

# Get all permutations  # https://www.geeksforgeeks.org/permutation-and-combination-in-python/
#  using library function 
from itertools import permutations # import lib
n = 7
lst = range(1,(n+1) ) # =  [1, 2, 3, 4, 5, 6, 7,..., n]
perm = permutations(lst) # all n! permutations are here


start_time = time.time()
_sum = 0
_new_power = 4.44
for p in list(perm):  # Loop over permutations
  inv_count = inversion(list(p)) # Calculate inversion number for "p"
  for i in range(1,(n+1)):
    _sum += (-1)**inv_count * (i**_new_power) *p[i-1]

print(_sum)

print('n=',n, time.time() - start_time  , 'seconds passed' )

For n= 9 it runs: 6.369966745376587 seconds passed on google's colab (it should be faster than notebook, but not much)