$$2\sum_{0\leq i <j \leq n}\sum_{k=1}^{j-i}\frac{(-1)^{i+j-1}}{j-i}{n \choose i}{n \choose j}\ln(n+i+k)$$ $$=2\sum_{k=1}^n\sum_{i=0}^{k-1} \sum_{j=k}^n {n \choose i}{n \choose j}\frac{(-1)^{i+j-1}}{j-i}\ln(n+k)$$
I have tried expanding both sums, also by induction. But I have not made it.
For the sake of space, let us denote $C(i, j, n) = \binom n i \binom n j \frac{(-1)^{i + j - 1}}{j - i}.$
First, one should notice that $0 \leq i < j \leq n$ if and only if $0 \leq i \leq j - 1$ and $i + 1 \leq j \leq n.$ Consequently, the sum $\sum_{0 \leq i < j \leq n}$ can be rewritten as $\sum_{j = 1}^n \sum_{i = 0}^{j - 1},$ and the entire first displayed sum can be rewritten as $$\sum_{0 \leq i < j \leq n} \sum_{k = 1}^{j - i} C(i, j, n) \ln(n + i + k) = \sum_{j = 1}^n \sum_{i = 0}^{j - 1} \sum_{k = 1}^{j - i} C(i, j, n) \ln(n + i + k).$$ Consider the set $S = \{(i, j, k) \,|\, 1 \leq k \leq j - i \text{ and } 0 \leq i \leq j - 1 \text{ and } 1 \leq j \leq n\}.$ Observe that the substitution $\ell = i + k$ with $n + \ell = n + i + k$ induces a bijection $$S \leftrightarrow S' = \{(i, j, \ell) \,|\, i + 1 \leq \ell \leq j \text{ and } 0 \leq i \leq j - 1 \text{ and } 1 \leq j \leq n\}.$$ From here, one can deduce (with a little effort) that $$\sum_{j = 1}^n \sum_{i = 0}^{j - 1} \sum_{k = 1}^{j - i} C(i, j, n) \ln(n + i + k) = \sum_{\ell = 1}^n \sum_{i = 0}^{\ell - 1} \sum_{j = \ell}^n C(i, j, n) \ln(n + \ell),$$ and this is equivalent to the second sum that you have in your displayed equation.