Can we find $x_{1}, x_{2}, ..., x_{n}$? Consider this.
$$x_{1}+x_{2}+x_{3}+....+x_{n}=a_{1}$$
$$x_{1}^2+x_{2}^2+x_{3}^2+....+x_{n}^2=a_{2}$$
$$x_{1}^4+x_{2}^4+x_{3}^4+....+x_{n}^4=a_{3}$$
$$x_{1}^8+x_{2}^8+x_{3}^8+....+x_{n}^8=a_{4}$$
$$.............................$$
$$x_{1}^{2^{n-1}}+x_{2}^{2^{n-1}}+x_{3}^{2^{n-1}}+....+x_{n}^{2^{n-1}}=a_{n}$$
Can we find $x_{1}, x_{2}, ..., x_{n}$ by knowing $a_{1}, a_{2}, ..., a_{n}$?

If you mean the diophantine-equation tag and the solutions are supposed to be integers, searching will be easy because high powers are spaced far apart.  Because of the symmetry you can insist that $x_1 \ge x_2 \ge \dots \ge x_n$.  You can focus just on the last equation $x_{1}^{2^{n-1}}+x_{2}^{2^{n-1}}+x_{3}^{2^{n-1}}+....+x_{n}^{2^{n-1}}=a_{n}$.  We have $x_{1}^{2^{n-1}} \le a_n \le nx_{1}^{2^{n-1}}$ or $(\frac{a_n}n)^{2^{1 \le -n}}  x_1 \le a_n^{2^{1-n}}$  Since $n^{2^{1-n}}$ is just a bit greater than $1$, this is a tight bound.  
For example, let $n=5$ and the final equation be $x_{1}^{16}+x_{2}^{16}+x_{3}^{16}+x_{4}^{16}+x_{5}^{16}=46512832447930819$  We get $9.955 \le a_1 \lt 11.008$, so $a_1$ must be $10$ or $11$.  If we take $a_1=10$, then $9.94 \lt a_2$, so $a_2=10$ as well.  It turns out we get driven to $10,10,10,10,10$, which is too large, or $10,10,10,10,9$, which is too small.  If we take $a_1=11$ things work better (because I made them).  We get $7.66 \lt a_2 \lt 8.35$, so $a_2=8$, then $7.47 \lt a_3 \lt 8.002,$ so $a_3=8$ and we find $a_4=5, a_5=3$ works.
