I've seen on the site and elsewhere calculations of determinants of the form $\det (\min(i,j))$ and $\det(\gcd(i,j))$.
There is a general formula in the following case: $\mathcal{O}$ is a finite partially-ordered set which is a lower semilattice, $a\colon \mathcal{O} \to R$ is a function from $\mathcal{O}$ to a commutative ring $R$.
Problem: calculate the determinant $$\det(a_{\inf(i,j)})_{i,j\in \mathcal{O}}$$
Note: the determinants of form $\det(a_{\min(i,j)})_{1\le i,j \le n}$ were calculated by George Boole (an example for $n=5$). His idea works in the general case.
The question is left for further reference purposes. Thank you for your interest!
$\bf{Added:}$ @amrsa: just pointed out that this setup works for any set $\mathcal{O}$ with a binary operation. We have a similarity with circulant matrices, or group determinants. In those cases we have some interesting formulas too.
$\bf{Added:}$
I will add a solution to the problem.
Let $\mathcal{O}$ a partially ordered set, $R$ an abelian group. Given a function $a\colon \mathcal{O}\to R$ there exists a (unique) function $b\colon \mathcal{O}\to R$ such that $$a(i) = \sum_{k\le i} b(k)$$
(see Mobius inversion formula)
The reason for the existence is the following: the linea map $b\mapsto a$ from $R^{\mathcal{O}}$ to itself is of the form $I + N$ where $N$ is a lower triangular matrix, so nilpotent. This implies $I+N$ is invertible, with the inverse given by a Neumann series $I-N + N^2 - \cdots$. So the Mobius function appears as the entries in this inverse matrix.
Now, consider the lower triangular matrix $(\delta_{i \ge k})_{i,k \in \mathcal{O}}$.
We have
$$a_i = \sum_{k \le i} b_i = \sum_k \delta_{i \le k} b_k$$
Now assume that $\mathcal{O}$ is also a lower-lattice, that is, every two elements $i$, $j$ have an infimum $\inf(i,j)$. We have
$$a_{\inf(i,j)} = \sum_{k \le \inf(i,j)} b_k = \sum_{k \le i \textrm{ and } k\le j} b_k = \sum \delta_{i\ge k} \cdot b_k \cdot \delta_{j\ge k}$$
We have gotten an Choleski decomposition of the symmetric matrix $A$. We conclude that $$\det A = \prod_{k \in \mathcal{O}} b_i$$
Note: the matrix $(\delta_{ik}) = (\delta_{i\ge k})$ is lower triangular unipotent since the nonzero entries are in the region $i \ge k$. Its transpose is upper triangular. This lower ( upper) thing works well for determinants.