The numbers n such that $\binom{n}{1}$ have $1$ prime factor (counted with multiplicity) are simply the primes. Therefore, for $k=1$ this gives the largest known prime, $n=2^{82589933}-1$. For $k=2$, the numbers n such that $\binom{n}{2}$ have $2$ prime factors (counted with multiplicity) are $2p$ such that $p$ and $2p-1$ are primes and the safe primes, primes $p$ such that $(p-1)/2$ is also prime. Since $2618163402417\cdot2^{1290001}-1$, the largest known prime $p$ such that $(p-1)/2$ is also prime, is greater than $7775705415\cdot2^{175116}+2=2p$, where $p$ is the largest known prime such that $2p-1$ is also prime, $2618163402417\cdot2^{1290001}-1$ is the largest known such $n$ for $k=2$. This will change from time to time.
What about $3\le k \le 32$?
For $3\le k\le 14$, I found in OEIS that $\binom{2918756139031688155200+k}{k}$, where $n=2918756139031688155200+k$, $k\le 14$, and $\binom{7272877497848202239}{k}$, where $n=7272877497848202239$, $k\le 14$, have k prime factors (counted with multiplicity). These are just the lower bounds for such $n$'s. Definitely these aren't the largest known $n$'s such that $\binom{n}{k}$ has $k$ prime factors (counted with multiplicity) for $3\le k \le6$.
Main problem: Find at least one positive integer $n\gt10^4$ such that $\binom{n}{k}$ has exactly $k$ prime factors (counting multiplicity) for each $15\le k\le 32$.