Let me formalize the problem and give some results.
A number $n$ is $y$-smooth (or $y$-friable) if all its prime factors $p$ are less than or equal to $y$. Let us set $$\Psi(x,y) :\# \{ n \le x: n \text{ is }y\text{-smooth}\},$$ which appeared in Sungjin Kim's comment. One can consider the quantity $$S_k(x,y) := \# \{ n_1,\ldots,n_k \le x: \gcd(n_1,n_2,\ldots,n_k)=1, \, n_i \text{ is }y\text{-smooth}\}.$$ The ratio $S_k(x,y)/\Psi(x,y)^k$ is exactly the probability that $n$ positive integers chosen uniformly at random from the set of $y$-smooth numbers up to $x$ have no (joint) non-trivial common factor.
If $n \le x$ then $n$ is $x$-smooth, so $S_k(x,x)/\Psi(x,x)^k = S_k(x,x)/\lfloor x \rfloor^k$ is the probability that $n$ positive integers $\le x$ chosen uniformly at random have no common factor, and this converges to $1/\zeta(k)$. You seem to ask about $S_k(x,y)/\Psi(x,y)^k$ when $y$ is fixed. This fraction is quite natural -- it represents the probability that a vector of $y$-smooth numbers and $L^1$-norm $\le x$ is visible from the origin.
We can use inclusion-exclusion to express $S_k(x,y)$ as $$\begin{align} S_k(x,y) &= \sum_{d \mid \prod_{p \le y}}\mu(d) \# \{ m_1,\ldots,m_k \le x/d: m_i \text{ is }y\text{-smooth}\}\\ &=\sum_{d\text{ is }y\text{-smooth}} \mu(d)\Psi(x/d,y)^k\end{align}$$ where $d$ stands for the common divisor of $k$ $y$-smooth numbers $n_1,\ldots,n_k$. So we need to understand $$ \sum_{d\text{ is }y\text{-smooth}} \mu(d)\frac{\Psi(x/d,y)^k}{\Psi(x,y)^k}.$$
A special case of a result of Ennola (1969, see discussion here) says that for fixed $y$, $\Psi(x,y) \sim C_y (\log x)^{\pi(y)}(1+O_y(1/\log x))$ where $\pi(y)$ is the number of primes up to $y$ and $C_y$ is an explicit positive constant. (This case is quite easy to prove, since $y$-smooth numbers correspond to lattice points in the $\pi(y)$-dimensional polytope $a_{i} \ge 0, \sum_{i=1}^{\pi(y)} a_{i} \log p_i \le \log x$).
Since $\log x \sim \log (x/d)$ for any fixed $d$, and since $\sum_{d\text{ is }y\text{-smooth}}\mu(d)=0$ for $y \ge 2$, this immediately implies that the above fraction is $O_y(1/\log x)$, so the probability tends to $0$. So most (in a limit sense) $k$-tuples of $y$-smooth integers have a common factor, as observed numerically by Stefan Kohl and Aaron Meyerowitz.
Below I discuss what happens if $y$ is allowed to grow with $x$. Luckily, the ratio $\Psi(x/d,y)/\Psi(x,y)$ (`local density') was studied extensively in the literature, namely in
- Adolf Hildebrand and Gérald Tenenbaum, "On integers free of large prime factors". Trans. Amer. Math. Soc. 296 (1986), no. 1, 265–290.
- Aleksandar Ivić and Gérald Tenenbaum, "Local densities over integers free of large prime factors". Quart. J. Math. Oxford Ser. (2) 37 (1986), no. 148, 401–417.
- Régis de la Bretèche and Gérald Tenenbaum, "Propriétés statistiques des entiers friables", Ramanujan J. 9 (2005), no. 1-2, 139–202.
- See also Theorem III.5.22 in Tenenbaum's book (3rd edition of English version).
Roughly speaking, in Theorem 3 of the Hildebrand--Tenenbaum paper, they show in certain ranges of parameters, $\Psi(x/d)/\Psi(x,y) \sim d^{-\alpha}$ (with reasonable error term) where $\alpha$ is a parameter depending only on $x$ and $y$ (arising as a saddle point of some generating function); $\alpha$ is studied in Theorem 2(i) and its proof. If $y\gg \log x$ it is close to $1-(\log \log x/\log y)$ (see also the Ivić--Tenenbaum for more precise results in the range $y \gg \log x$), while for small $y$ it is close to $y/(\log x \log y)$.
Carefully using these results leads to a result of the shape $$(\star)\, \sum_{d\text{ is }y\text{-smooth}} \mu(d)\frac{\Psi(x/d,y)^k}{\Psi(x,y)^k} \sim \sum_{d\text{ is }y\text{-smooth}} \mu(d)d^{-k\alpha} = \prod_{p \le y} \left(1-p^{-k\alpha}\right)$$ for some range of $y$ and $x$, where $\alpha$ is the above-mentioned constant. If, say, $y=(\log x)^C$ for some $C \ge 1+\varepsilon$, this $\alpha=1-1/C+o(1)$ and $\Psi(x,y)=x^{1-1/A+o(1)}$. A short computation shows that if $y \ge (\log x)^{1+1/(k-1)+\varepsilon}$ then this probability is bounded away from $0$, while if $y \le (\log x)^{1+1/(k-1)-\varepsilon}$ it goes to $0$.
I'll end by mentioning that in the La Bretèche--Tenenbaum paper they study the number of $y$-smooth numbers up to $x$ that are coprime to a given integer $m$, obtaining results uniform in $m$. This can be expressed as $\sum_{d \mid m}\mu(d) \Psi(x/d,y)$, which is similar to the quantity mentioned above, so maybe the best results are obtained using the ideas in that work. In particular, they show that $$\Psi(x/d,y) \ll \Psi(x,y)d^{-\alpha}$$ uniformly for $x \ge y \ge 2$, $1 \le d \le x$, which is powerful when trying to truncate the sum over $d$ in $(\star)$.
I'll return to this and give a formal argument for $(\star)$ when $y$ is not too small.