What is the set of all minimal 3CNF formulas?
The number of minimal formula is finite.  Indeed, if a formula is minimal, then no clause can be repeated.  Now there are $(2n)^3=8n^3$ possible clauses, so there are at most $2^{8n^3}$ different formulas with no repeated clauses.  Thus there are at most $2^{8n^3}$ minimal formula, which is finite.

I don't know of any clean way to count the number of minimal formula; probably there is no simple formula for it.  You could try computing the number for $n=1,2,3$ by brute-force, then looking up the resulting sequence in OEIS, but I don't expect you to find any simple expression for this number.

What you are thinking is wrong.  A minimal, unsatisfiable formula can have more than 8 clauses.