It is easy to determine that $-4|x-1|$ will reach its maximum value of $0$ when $x=1$. Meanwhile, exponential function is monotonically increasing when its base is greater than $1$. Therefore, $3^{-|\log_2x|}$ will reach its maximum when $-|\log_2x|$ is the largest, or $x=1$. Hence, the $\max f(x) = f(1) = 1$. Obviously, the minimal value of $f(x)$ is $-\infty$ and this function is continuous on its entire domain which means its range is $\boxed{(-\infty, 1]}$.