This is not an answer, either. Just some attempts to attack the problem.

Let $r(n)$ be the square-free part of $(2, 3)_n$, i.e. $r(n)$ is square-free and we have $(2, 3)_n = r(n) B^2$ for some integer $B$. We have the following table of the values of $r(n)$:

```
n r(n)
11 -2
23 2
37 1
47 -6
59 -2
61 1
71 -2
73 -5
83 6
97 -3
107 -6
109 -1
131 6
157 -7
167 2
179 -6
181 7
191 3
193 -1
227 -2
229 -5
239 2
241 7
251 -10
253 -1
263 -1
277 -1
313 -13
337 -1
347 2
349 -7
359 -10
373 -7
383 -1
397 5
```

Note the composite number $253 = 11 \times 23$, with $r(253) = r(11)r(23)$. This seems to be true in general (identities in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$):

```
n r(n)
11*23 -1 = r(11) * r(23)
11*37 -2 = r(11) * r(37)
11*47 3 = r(11) * r(47)
11*59 1 = r(11) * r(59)
23*37 2 = r(23) * r(37)
23*47 -3 = r(23) * r(47)
```

This observation might be proved by using Chinese remainder theorem to rewrite the matrix as some sort of tensor product (not sure about the details, but seems doable). For non-square-free $n$, however, the determinant seems always zero.

This suggests that we could concentrate on the case where $n = p$ is a prime number.

In this case, the original conjecture (i.e. $p^2$ divides $(2, 3)_p$) is implied by the following two statements:

$p$ divides $(2, 3)_p$;

$p$ does not divide $r(p)$.

Now 1. might be easier (than the original problem) to prove, since we can then do calculations in $\mathbb{F}_p$, and write the Jacobi symbol as $x^{p-1}$. This is again just a very rough idea.

As to 2., the thing to notice is that, although $(2, 3)_n$ is extremely large, the numbers $r(n)$ are surprisingly small. I don't see any explanation to this phenomenon.

Concerning generalizations of these statements:

Statement 2. seems to be quite general. For almost arbitrary choice of parameters $(c, d)$, one always observes very small values of $r(n)$.

Statement 1., however, depends strongly on the choices of $(c, d)$. So far I only see this phenomenon occuring for the following choices:

```
"good" (c, d)
(2, 3)
(4, 12)
(6, 27)
(8, 48)
......
(6, 15)
(12, 60)
(18, 135)
(24, 240)
......
```

Of course, a pair $(c, d)$ is not quite different from $(ac, a^2d)$, since the columns are just permuted, and the determinants only possibly differ by sign. So essentially there are only two "primitive" cases: $(c, d) = (2, 3), (6, 15)$. These two are essentially different, as can be seen from the difference between the $r(n)$ values.

I have no idea how to find more examples of "good" pairs $(c, d)$.

UPDATE:

I think it might be more natural to look at the $n \times n$ matrix, by adding the indices with $i = 0$ or $j = 0$. Let us call the determinant of this larger matrix $[2, 3]_n$.

It appears that in most cases $[2, 3]_n$ is a multiple of $(2, 3)_n$. In particular, when $n = p$ is a prime number for which $(2, 3)_p$ is non-zero, then we have $[2, 3]_p = \frac{p - 1}{2}(2, 3)_p$. When $n = pq$ is a product of two primes for which $(2, 3)_n$ is non-zero, then we have $[2, 3]_n = \frac{p-1}{2}\frac{q-1}{2}(2,3)_n$.

The advantage of considering $[2, 3]_n$ is the following obvious formula:
$$ [2, 3]_{nm} = [2, 3]_n^m [2, 3]_m^n, $$
whenever $n, m$ are coprime. This then reduces the statements above to the case of prime numbers.