How can a 32bit CPU have an addressable memory size of 16TB? Frequently, on a 32-bit CPU, each page-table entry is 4 bytes long, but that
size can vary as well.

A 32-bit entry can point to one of $2^{32}$ physical page frames.

If frame size is 4 KB (212), then a system with 4-byte entries can address $2^{44}$ bytes
(or 16 TB) of physical memory.

The above statement is taken from the book "Operating System Principles" by Galvin.

If all 32 bits in a 32-bit CPU are used to refer to pages , then we can have $2^{32}$ pages. But then no more bits will be left to point to memory inside a page of size $2^{12}$ bits since all 32-bits have been used up.

How can we thus say that $2^{44}$ bytes are addressable?
The $32$ bit page frame address acts as a base address and will be typically stored in an index register.

An individual machine code instruction (e.g. a branch instruction) will then contain a $12$ byte bit offset. The offset is added to the base address to create the complete $44$ byte bit address.