What is the probability that $n$ integers chosen at random are coprime?

It is extremely well-known that the probability of any two random integers being relatively prime is $\zeta(2)^{-1}$ (see here).

From An Introduction to Analytic Number Theory by Apostol, the proof involves counting lattice points and finding the limit $$\lim_{r\to\infty}\frac{N'(r)}{N(r)}$$ where $N(r)$ is the number of lattice points in the square governed by $|x|\le r$ and $|y|\le r$, and $N'(r)$ is the number of lattice points visible from the origin.

For some background, the term "visible" is defined as follows:

Two lattice points $P$ and $Q$ are said to be visible if the line joining the two does not go through any other lattice points.

Now what if $n\neq2$; that is,

What is the limit of the probability that $n$ integers chosen at random in the interval $[1,N]$ are coprime as $N\to\infty$ with $n>2$?

Let's consider the simplest case: $n=3$. We imitate the proof for when $n=2$.

Firstly, an extension to Thm 3.8 can be easily proven.

Theorem: Two lattice points $(a_1,\cdots,a_n)$ and $(b_1,\cdots,b_n)$ are visible iff $a_1-b_1$, $a_2-b_2$ up to $a_n-b_n$ are relatively prime.

The $24$ lattice points nearest the origin are all visible from the origin - there are $24$ points "surrounding" the origin of unit distance. By symmetry, we see that $N'(r)$ is equal to $24$, plus $24$ times the number of visible points in the region $$\{(x,y,z):2\le x\le r, ???\}$$ We cannot use $1\le y\le x$ since the gradient of the line joining the origin and $(r,r,r)$ is no longer $1$. Of course, we can try to use this, but I feel that this makes it more complicated than it should be.

So how should I continue? Is there an alternative method? And what would be the general approach for large $n$, ie. is there an expression (in terms of $n$) that finds the probability that $n$ integers chosen at random are coprime?

Let us clarify that from now on the meaning of random integer is "consider what happens for random elements of $[1,N]$ picked according to a uniform distribution, then let $N\to +\infty$".
For a fixed prime $p$ and three random integers $a,b,c$, the probability that $p$ does not divide more than one element among $\{a,b,c\}$ is $\left(\frac{p-1}{p}\right)^3 + \frac{3(p-1)^2}{p^3}$. It follows that the probability that three random integers are mutually coprime is
$$\prod_{p}\left(1-\frac{3p-2}{p^3}\right)\approx 28.67\%$$ and similarly the probability that four random integers are mutually coprime is $$\prod_{p}\frac{(p-1)^3(p+3)}{p^4}\approx 11.48\%$$ and the probability that five random integers are mutually coprime is $$\prod_{p}\frac{(p-1)^4(p+4)}{p^5}\approx 4.09\%$$ etcetera. Asymptotically, the probability that $m$ random integers are mutually coprime behaves like $K e^{-J m}$ for some positive constants $J,K$.