If $n$ is even, then the answer is "yes". Take the direct product of a simple (infinite) group and ${\mathbb Z}/2^j{\mathbb Z}$. Every normal subgroup either is inside the finite cyclic group or contains the simple group. Total number is twice the number of normal subgroups of the cyclic group.If $n$ is odd, you would need to take a cyclic central extension of a simple group. That is also possible (the simple group can be, say, the Tarski monster, see our paper with Olshanskii and Osin on Lacunary hyperbolic groups in the arXiv).