The number of positive squarefree integers $i \le n$ is given by: $$C(n)=\sum_{k=1}^{\lfloor\sqrt{n}\rfloor}\mu(k)\left\lfloor\frac{n}{k^{2}}\right\rfloor.$$ The number of positive integers $i\le n$ coprime to the first $k$ prime numbers admits the recurrence relation: $$\phi(n, k) = \lfloor n / p_k \rfloor - \phi(\lfloor n/p_k \rfloor, k-1) + \phi(n, k-1)$$
But I haven't managed to find any literature regarding the subject of numbers that are both squarefree AND coprime to the first $k$ prime numbers, with the exception of odd/even such integers ($k=1, p = 2$). Let $P_k = p_1\cdot p_2 \cdot...\cdot p_k$ be the primorial of the first $k$ primes. If we define $$f(n,k) = \sum_{i = 1 \atop gcd(i,P_k)=1}^{n}|\mu(i)|$$
to be the count of squarefree integers coprime to the first $k$ primes, then does this formula admit any elegant representations? Is there any way to efficiently exactly count (not an approximation) such numbers mathematically or at least compute this count efficiently?
To approach your problem, I would just write down your formula in terms of images and inverse images of the involved functions.
So take $|\mu|(n)$ to be the absolute value of $\mu(n)$ for each $n \in \Bbb{N}$.
Then:
$$ f(n,k) = \sum_{a=1}^n|\mu|(a \in \gcd^{-1}( p_1\cdots p_k,\cdot)(1)) $$
Where, if you look at $\gcd(p_1\cdots p_k, n) = g(n)$ as a multiplicative function, then we're interested in $\ker g$.
Thus $f(n,k) = |[1,..,n]\cap \ker |\mu| \cap \ker g|$,
$|\mu|$ also being a multiplicative function.
This is about as elegant a formula that I can come up with.
Note that $\ker g$ here is defined here as $\ker g = \{ n \in \Bbb{N}: g(n) = 1\}$, but the set $\ker |\mu|, \ker g$ are not monoids. They do however satisfy: if $a, b \in \ker g: \gcd(a,b) = 1 \implies ab \in \ker g$.
This means that the limit supremum exists for $a_n = f(n,k)/n$, i.e. $\lim\sup_{n \to \infty} a_n = \overline{d}(A)$, where $A = \ker g \cap \ker |\mu|$.
See natural density examples. Since the bound is from above, we must have: $\overline{d}(A) \leq \lim_{n \to \infty} \dfrac{|[1..n] \cap \ker |\mu||}{n} = 6/\pi^2$.
And $\underline{d}(A) \geq 0$.
I thought that I'd mention natural density because the problem naturally takes that form (just add a denominator $n$ and take lim sup/inf).
We also have a related formula:
$$\ker |\mu| \cap \ker g = \ker (|\mu| \cdot g) = \{ n \in \Bbb{N} : |\mu(n)|g(n) = 1 \}$$, where $|\mu| \cdot g$ is also a multiplicative function, i.e. for all $a,b \in \Bbb{N}$ such that $\gcd(a,b) = 1$, you have that $|\mu|(ab)g(ab) = (|\mu|(a)g(a))(|\mu|(b)g(b))$.