Partial answer: $f(5) \ge 11$.
A short proof shows that $f(5) \ge 10$. To be 5-universal (i.e. universal for all partial orders of 5 elements), our poset must contain a 5-chain. Also it must contain two incomparable 2-chains, only one of which can be in the 5-chain. Also it must contain 5 incomparable elements (only two of which could be in the two chains). So at least 5+2+1+1+1 = 10 elements. I believe this is essentially the kind of lower-bound argument that was mentioned in the earlier question. This "multiple chains" argument says nothing about branching structures in the 5-posets, so perhaps one could consider them and work out an improved lower bound.
A really brute-force SageMath code (see below) tries all 10-posets (about 2.6 million) in about 15 hours, and finds no 5-universal posets, so it must be at least 11 elements.
The code does not work well for 11-posets. There are not that many (less than 47 million), but at least in SageMath 8.8, Posets() generates the posets rather slowly. Most (~90%) of the time is actually spent in poset generation, not in checking for universality. Also I don't know of any public databases of all 11-posets that one could use here. However, there is an old enhancement request for faster poset generation in SageMath, and the Brinkmann & McKay code is embedded there! It should be lightning fast compared to Posets(), so then the bottleneck probably moves to the universality check.
# Find an u-poset that contains all n-posets as induced posets.
def find_universal_poset(n,u):
PP = list(Posets(n))
for U in Posets(u):
ok = True
for P in PP:
if not U.has_isomorphic_subposet(P):
ok = False
break
if ok:
return U
return None