Since $f(x)$ is an even function and it's monotonically decreasing on $[0, +\infty)$, the inequality $f(2^{x}) > f\left( -\frac{1}{2} \right)$ can be rewritten as $f(2^{x}) > f\left( \frac{1}{2} \right)$ due to the evenness of the function, which implies that $f(x)$ has the same value at $-\frac{1}{2}$ and $\frac{1}{2}$.

Now, because $f(x)$ is monotonically decreasing on $[0, +\infty)$, for the inequality $f(2^{x}) > f\left( \frac{1}{2} \right)$ to hold, the argument $2^{x}$ must be less than $\frac{1}{2}$.

Let's analyze the inequality $2^{x} < \frac{1}{2}$:
We know that $2^{-1} = \frac{1}{2}$, thus we can equate the exponents since the bases are the same:
$$x < -1$$

Hence, the range of $x$ that satisfies the inequality is $(-\infty, -1)$.

Therefore, the correct answer is $\boxed{C}$.