How many numbers $ N \le 10^{10}$ are the product of $3$ distinct primes?

You'll need a list of primes out to around $71{,}000$. This is because if $n\lt10^{10}$ is a product of three distinct primes, say $n=pqr$ with $p\lt q\lt r$, then $q\lt\sqrt{10^{10}/p}\le10^5/\sqrt2\approx70{,}710.678$. Note also that $p\lt\sqrt[3]{10^{10}}\approx2154.43$, so there is a fairly straightforward brute-force algorithm that seems practical, at least for the specific number, $10^{10}$:

For each prime $p$ between $2$ and $2154$, consider the primes $q$ between $p$ and $10^5/\sqrt p$. For each of these, compute $\pi(\lfloor10^{10}/pq\rfloor)-\pi(q)$, which counts the number of available primes $r$ that are greater than $q$. There are only a few hundred primes $p$ and, for most of these, only a few thousand primes $q$, so the total number of times the $\pi$ function needs to be evaluated is almost certainly less than a million, which seems reasonable to ask a computer to do. A more careful analysis could possibly shave off another order of magnitude or so.

In sum, we can write the function the OP is after as

$$\pi_3(N)=\sum_{2\le p\le\sqrt[3]{N}}\left(\sum_{p\lt q\le\sqrt{N/p}}\left(\pi(\lfloor N/pq\rfloor)-\pi(q) \right)\right)$$

where the nested sums are understood to run over primes in the given ranges. The total number of summands is (substantially) less than $N^{1/3}N^{1/2}=N^{5/6}$. I'd be happy to see an approach that does better.