Assuming the two subsets are selected independently, the probability in question is $$p_n=a_n/2^{2n},$$ where $a_n$ the sum of the squares of the coefficients in the polynomial $$\prod_{k=1}^n (1+x^k).$$ The sequence $(a_n)$ is the sequence A047653, and its asymptotics, given on that page, is $$a_n\sim\frac{\sqrt{3/\pi}\,4^n}{n^{3/2}}$$ (as $n\to\infty$). So, $$p_n\sim\frac{\sqrt{3/\pi}}{n^{3/2}}.$$

Assuming the two subsets are selected independently, the probability in question is $$p_n=a_n/2^{2n},$$ where $a_n$ the sum of the squares of the coefficients in the polynomial $$\prod_{k=1}^n (1+x^k).$$ The sequence $(a_n)$ is the sequence A047653, and its asymptotics, given on that page, is $$a_n\sim\frac{\sqrt{3/\pi}\,4^n}{n^{3/2}} \tag{1}$$ (as $n\to\infty$). So, $$p_n\sim\frac{\sqrt{3/\pi}}{n^{3/2}}.\tag{2}$$

For (1), A047653 refers to Vaclav Kotesovec, where I have been unable to find a link/reference to (1). However, the equivalent relation (2) was actually proved by van Lint. (I found a link to van Lint on the related OEIS page A000980.)

Assuming the two subsets are selected independently, the probability in question is $$p_n=s_n/2^{2n},$$ where $a_n$ the sum of the squares of the coefficients in the polynomial $$\prod_{k=1}^n (1+x^k).$$ The sequence $(a_n)$ is the sequence A047653, and its asymptotics, given on that page, is $$a_n\sim\frac{\sqrt{3/\pi}\,4^n}{n^{3/2}}$$ (as $n\to\infty$). So, $$p_n\sim\frac{\sqrt{3/\pi}}{n^{3/2}}.$$

Assuming the two subsets are selected independently, the probability in question is $$p_n=a_n/2^{2n},$$ where $a_n$ the sum of the squares of the coefficients in the polynomial $$\prod_{k=1}^n (1+x^k).$$ The sequence $(a_n)$ is the sequence A047653, and its asymptotics, given on that page, is $$a_n\sim\frac{\sqrt{3/\pi}\,4^n}{n^{3/2}}$$ (as $n\to\infty$). So, $$p_n\sim\frac{\sqrt{3/\pi}}{n^{3/2}}.$$