Asymptotic behavior of number of triples $i,j,k\le n$ with pairwise bounded least common multiples each $\le n$.

In this answer I'll prove that:

Proposition 1: We have the upper bound $$f(n)\leq 8n^{\frac{3}{2}}\zeta\left(\frac{3}{2}\right)^{4}$$

In principle it is possible to proceed from equation $(1)$ below and obtain an exact asymptotic for $f(n)$.

Proof. To better understand the triple sum, lets examine the summands, and attempt to rearrange the order of summation. For a given triple $i,j,k$, let $r=(i,j,k)$ denote the greatest common divisor among $i,j,k$. Let $\alpha,\beta,\gamma$ denote the leftover pairwise greatest common divisors: $$\alpha=\left(\frac{i}{r},\frac{j}{r}\right),\ \ \ \beta=\left(\frac{j}{r},\frac{k}{r}\right),\ \ \ \gamma=\left(\frac{k}{r},\frac{i}{r}\right).$$ Then $\alpha,\beta,\gamma$ will be pairwise relatively prime, and we may write $$i=r\alpha\gamma i',\ \ \ j=r\alpha\beta j',\ \ \ k=r\beta\gamma k'$$ where $i',j',k'$ are pairwise relatively prime. This will be a valid triple so long as each of $i'j'$, $j'k'$, and $k'i'$ are each at most $\frac{n}{r\alpha\beta\gamma}$. This is because $$\text{lcm}(i,j)=r\alpha\beta\gamma i'j'\leq n.$$ Thus rearranging the sum, we have that $$f(n)=\sum_{r\leq n}\sum_{\substack{\alpha,\beta,\gamma\leq\frac{n}{r}\\ \alpha,\beta,\gamma\text{ pairwise}\\ \text{relatively prime} } }\sum_{\substack{ij,jk,ki\leq\frac{n}{r\alpha\beta\gamma}\\ i,j,k\text{ pairwise}\\ \text{relatively prime}} }1\ \ \ \ \ \ \ \ \ \ (1)$$ It follows that we may upper bound $f(n)$ by $$f(n)\leq \sum_{r\leq n}\sum_{\alpha,\beta,\gamma\leq\frac{n}{r}}\sum_{ij,jk,ki\leq\frac{n}{r\alpha\beta\gamma}}1.\ \ \ \ \ \ \ \ \ \ (2)$$ Let's proceed by bounding the inner sum $$\sum_{ij,jk,ki\leq x}1.$$ If $i>\sqrt{x}$, then we must have both $j,k\leq\sqrt{x}$ since $ij\leq x$ and $ik\leq x$. By symmetry it follows that $$\sum_{ij,jk,ki\leq x}1 =\sum_{i,j,k\leq\sqrt{x}}1+3\sum_{\sqrt{x}<i<x}\sum_{j,k\leq\frac{x}{i}}1$$ $$=\left[\sqrt{x}\right]^{3}+3\sum_{\sqrt{x}<i<x}\left[\frac{x}{i}\right]^{2}$$ where $\left[x\right]$ denotes the floor of $x$. Since $[x]=x-\{x\}\leq x$, the above is at most $$x^{\frac{3}{2}}+3x^{2}\sum_{\sqrt{x}<i\leq x}\frac{1}{i^{2}}.$$ For $x<2$, the rightmost sum is $0$. For $x\geq2$, by considering the area under the curve $1/y^{2}$, the sum over $i$ can be bounded by the integral $$\int_{\sqrt{x}-1}^{x-1}\frac{1}{y^{2}}dy=\frac{1}{\sqrt{x}-1}-\frac{1}{x-1}=\frac{1}{\sqrt{x}}\cdot\left(\frac{x}{x-1}\right)\leq\frac{2}{\sqrt{x}}.$$ Hence $$\sum_{ij,jk,ki\leq x}1\leq8x^{\frac{3}{2}}.$$ It follows that $$f(n)\leq\sum_{r\leq n}\sum_{\alpha,\beta,\gamma\leq\frac{n}{r}}8\frac{n^{\frac{3}{2}}}{r^{\frac{3}{2}}\alpha^{\frac{3}{2}}\beta^{\frac{3}{2}}\gamma^{\frac{3}{2}}}$$ $$\leq8n^{\frac{3}{2}}\sum_{r}\frac{1}{r^{\frac{3}{2}}}\sum_{\alpha,\beta,\gamma}\frac{1}{\alpha^{\frac{3}{2}}\beta^{\frac{3}{2}}\gamma^{\frac{3}{2}}}$$ $$\leq8n^{\frac{3}{2}}\zeta\left(\frac{3}{2}\right)^{4}.$$