The limit $\lim_{n \to \infty} \frac{\rho(A)}{n^2}$ is equal to the largest positive (real) solution to $$4x\cos^4\bigl(\frac{1}{\sqrt{2x}}\bigr)=\sqrt{\frac{x}{2}}\sin\bigl(\sqrt{\frac{2}{x}}\bigr)\Bigl(2-\sqrt{2x}\sin(\sqrt{\frac{2}{x}})\Bigr).$$

Indeed, in the limit, an eigenvector $\phi : [0,1] \to \mathbb{C}$ has eigenvalue $\lambda$ if $$\int_0^1 \Bigl(1-|x-y|\Bigr)\phi(y)dy = \lambda \phi(x)$$ for each $x \in [0,1]$.

Assuming we can write $\phi(x) = \sum_{n=0}^\infty a_n x^n$ as a power series, we obtain the equations $$\lambda a_0 = \sum_{n=0}^\infty \frac{1}{(n+1)(n+2)}a_n,$$ $$\lambda a_1 = \sum_{n=0}^\infty \frac{1}{n+1}a_n,$$ $$\lambda a_n = -\frac{2}{(n-1)n} a_{n-2},$$ the last being for $n \ge 2$. One derives from the last equation that $a_{2k} = \frac{(-2\lambda^{-1})^k}{(2k)!}a_0$ and $a_{2k+1} = \frac{(-2\lambda^{-1})^k}{(2k+1)!}a_1$ for $k \ge 0$. Substituting these values into the first two equations gives $$\lambda a_0 = \lambda \sin^2(\frac{1}{\sqrt{2\lambda}}) a_0 + \frac{2-\sqrt{2\lambda}\sin(\sqrt{\frac{2}{\lambda}})}{4}\lambda a_1$$ $$\lambda a_1 = \sqrt{\frac{\lambda}{2}}\sin(\sqrt{\frac{2}{\lambda}})a_0 + \lambda\sin^2(\frac{1}{\sqrt{2\lambda}}) a_1.$$ Simplifying and then dividing the equations gives $$\frac{\cos^2(\frac{1}{\sqrt{2\lambda}})}{\sqrt{\frac{\lambda}{2}}\sin(\sqrt{\frac{2}{\lambda}})} = \frac{2-\sqrt{2\lambda}\sin(\sqrt{\frac{2}{\lambda}})}{4\lambda \cos^2(\frac{1}{\sqrt{2\lambda}})}.$$

Because the $a_n$'s decay very rapidly, the largest positive (real) solution to the above will yield a valid function $\phi$. So all that remains is to argue that any eigenvector $\phi$ has a power series expansion. Maybe this is standard?

The limit $\lim_{n \to \infty} \frac{\rho(A)}{n^2}$ is equal to $2/y^2$, where $y$ is the smallest positive real solution to $$2\cos y + 2 = y\sin y.$$

Indeed, in the limit, an eigenvector $\phi : [0,1] \to \mathbb{C}$ has eigenvalue $\lambda$ if $$\int_0^1 \Bigl(1-|x-y|\Bigr)\phi(y)dy = \lambda \phi(x) \tag{1}$$ for each $x \in [0,1]$.

By $(1)$, $\phi$ is regular enough so that we can write $\phi(x) = \sum_{n=0}^\infty a_n x^n$ as a power series, thereby obtaining, from $(1)$, the equations $$\lambda a_0 = \sum_{n=0}^\infty \frac{1}{(n+1)(n+2)}a_n,$$ $$\lambda a_1 = \sum_{n=0}^\infty \frac{1}{n+1}a_n,$$ $$\lambda a_n = -\frac{2}{(n-1)n} a_{n-2},$$ the last being for $n \ge 2$. One derives from the last equation that $a_{2k} = \frac{(-2\lambda^{-1})^k}{(2k)!}a_0$ and $a_{2k+1} = \frac{(-2\lambda^{-1})^k}{(2k+1)!}a_1$ for $k \ge 0$. Substituting these values into the first two equations gives $$\lambda a_0 = \lambda \sin^2(\frac{1}{\sqrt{2\lambda}}) a_0 + \frac{2-\sqrt{2\lambda}\sin(\sqrt{\frac{2}{\lambda}})}{4}\lambda a_1$$ $$\lambda a_1 = \sqrt{\frac{\lambda}{2}}\sin(\sqrt{\frac{2}{\lambda}})a_0 + \lambda\sin^2(\frac{1}{\sqrt{2\lambda}}) a_1.$$ Simplifying and then dividing the equations gives $$\frac{\cos^2(\frac{1}{\sqrt{2\lambda}})}{\sqrt{\frac{\lambda}{2}}\sin(\sqrt{\frac{2}{\lambda}})} = \frac{2-\sqrt{2\lambda}\sin(\sqrt{\frac{2}{\lambda}})}{4\lambda \cos^2(\frac{1}{\sqrt{2\lambda}})}.$$

Now just let $y = \sqrt{2/\lambda}$ and observe $4\cos^4 (y/2) = (\sin y)(y-\sin y)$ is equivalent to $2\cos y + 2 = y\sin y$.

[Edit: I now just realized $(1)$ implies $\lambda \phi''(x) = -2\phi(x)$, yielding $\phi(x) = C_1\sin\Bigl(\sqrt{\frac{2}{\lambda}}x\Bigr)+C_2\cos\Bigl(\sqrt{\frac{2}{\lambda}}x\Bigr)$.]