Any group of order $n$ is cyclic if and only if $n$ is square-free and no prime factor of $n$ is congruent to one modulo any other prime factor of $n$.* Thus, the smallest such $n$ divisible by two primes is $3\cdot5=15$, the smallest divisible by three primes is $255=3\cdot5\cdot17$, etc. Dirichlet's theorem on primes in arithmetic progression guarantees that given any such $n$ we can always find a multiple of $n$ containing an additional prime factor, that also has this property. (Just take the sequences with steps of size $n$ starting with a number that is coprime to $n$ and not congruent to $1$ mod $n$.)
This there are such numbers $n$ with as many prime factors as you please. And it isn't too difficult to construct them by the greedy method of looking for the next prime that isn't $1$ mod any primes already selected. But it seems unlikely this will always give the smallest $n$ corresponding to a given number of prime factors. Is there a way (more elegant than sheer brute force) to determine for each $k$ the smallest $n$ that is a square-free product of $k$ distinct primes none of which is congruent to $1$ mod one of the others?
* If you are interested in the group theory argument, square-free is an obvious necessity, and if $p$ divides $q-1$, then there is a non-abelian semi-direct product. The converse is a consequence of the counting portion of Philip Hall's theorems..