The conjecture for $n=4$ is correct.  For general $n$ and any $x_1,\ldots,x_n$,
$$
f(p) := \sum_{i=1}^{n}\dfrac{x^p_{i}}{\prod_{j\neq i}(x_{i}-x_{j})}
$$
is positive for $p > n-2$ and switches sign at $p=0,1,2,\ldots,n-2$
and nowhere else.  For example:   
If $n=5$ then $f(p) \geq 0$ iff 
$p \geq 3$ or $1 \leq p \leq 2$ (or $p \leq 0$);  
if $n=6$ then $f(p) \geq 0$ for $p \in [0,1]$, $[2,3]$, or $[4,\infty)$;   
if $n=7$ then $f(p) \geq 0$ for 
  $p \in (-\infty,0]$, $[1,2]$, $[3,4]$, or $[5,\infty)$;  
etc.

The key fact is the following exponential analogue of "Descartes' law of sines":

**Proposition.** *Let $x_1,\ldots,x_n$ be pairwise distinct 
positive real numbers, and $c_i$ any real numbers not all zero.
Then the function $f(p) := \sum_{i=1}^n c_i^{\phantom.} x_i^p$
has at most $n-1$ real zeros, counted with multiplicity.*

*Proof*: Induction on $n$.  For $n=1$ the function $f$ is a nonzero multiple of
$x_1^p$, which is never zero; this establishes the base case.
Now let $n>1$, and assume we have proved the case $n-1$ of the Proposition.
Let $f_1(p) = x_n^{-p} f(p) = \sum_{i=1}^n c_i (x_i/x_n)^p$, which is 
of the same form as $f$ with $x_n=1$ and has the same zeros and multiplicities.
Then the inductive hypothesis applies to the derivative 
$$
f'_1(p) = \sum_{i=1}^{n-1} (c_i \log (x_i/x_n)) \, (x_i/x_n)^p;
$$
so $f'_1$ has at most $n-2$ zeros, counted with multiplicity.
By Rolle's Theorem it follows that $f_1$ has at most $n-1$ zeros,
counted with multiplicity.  This completes the induction step and the proof.
**QED**

It remains to check that our $f$ vanishes at $p=0,1,\ldots,n-2$;
then we can use our Proposition to deduce that each of these $n-1$ zeros
is simple and there are no others.  This can be done by recognizing
$f(p)$ as the sum of the residues of the differential 
$x^p \, dx \left/ \prod_{j=1}^n (x-x_j) \right.$
(which is regular at $x=\infty$ for integers $p \leq n-2$).
Alternatively we can relate $f(p)$ to
a Schur-Vandermonde determinant with a repeated row.