Assume that $a_n> 0$ for every positive integer $n$; otherwise the limit may not exist. Set $z_n:=n^x\,a_n$ like Martin R recommended. Thus,
$$n^x\,\left(\prod_{i=1}^n\,a_i\right)^{\frac{1}{n}}=\frac{n^x}{\left(\prod\limits_{i=1}^n\,i^x\right)^{\frac1n}}\,\left(\prod\limits_{i=1}^n\,z_i\right)^{\frac1n}=\left(\frac{n^n}{n!}\right)^{\frac{x}{n}}\,\left(\prod_{i=1}^n\,z_i\right)^{\frac1n}\,.$$
Now, since $\displaystyle\lim_{n\to\infty}\,z_n=a$, we have $\displaystyle\lim_{n\to\infty}\,\left(\prod_{i=1}^n\,z_i\right)^{\frac1n}=a$ (see Martin R's link in the comments above). Furthermore, Stirling's approximation $n!\approx \sqrt{2\pi n}\left(\frac{n}{\text{e}}\right)^n$ implies that
$$\lim_{n\to\infty}\,\left(\frac{n^n}{n!}\right)^{\frac{x}{n}}=\lim_{n\to\infty}\,\left(\frac{\text{e}^n}{\sqrt{2\pi n}}\right)^{\frac{x}{n}}=\exp(x)\,.$$ Consequently, $$\lim_{n\to\infty}\,n^x\,\left(\prod_{i=1}^n\,a_i\right)^{\frac{1}{n}}=a\,\exp(x)\,.$$
Let us define
$$\Sigma_{\infty,n} = \lim_{m\to\infty} \Sigma_{m,n}$$
It can be proven using simple induction that $\Sigma_{m,1}=1$ for all $m\in\mathbb N$, hence we have that $\Sigma_{
\infty,1}=1$. Now, from your recurrence relation, taking the limit of both sides as $m\to\infty$, we have
$$\Sigma_{\infty,n}=\sum_{k=1}^n \frac{\Sigma_{\infty,k}}{k^2}=\Sigma_{\infty,n-1}+\frac{\Sigma_{\infty, n}}{n^2}$$
which may be solved for $\Sigma_{\infty,n}$, yielding
$$\Sigma_{\infty, n}=\frac{\Sigma_{\infty, n-1}}{1-\frac{1}{n^2}}$$
Now, since we already have that $\Sigma_{\infty, 1}$, we obtain the following formula by induction:
$$\Sigma_{\infty, n} = \prod_{k=2}^n \Big(1-\frac{1}{k^2}\Big)^{-1}$$
which gives us the desired result:
$$\color{green}{\Sigma_{\infty,\infty} = \prod_{k=2}^\infty \Big(1-\frac{1}{k^2}\Big)^{-1} = 2}$$
which is supported by your numerical evidence, showing values $\approx 1.9$.
Best Answer
Let $$\begin{split} S_{k,p} &= \sum_{i=0}^p\binom{p}{i}\frac{(-1)^i}{i+k}\\ &= \sum_{i=0}^p\binom{p}{i}(-1)^i\int_0^1 x^{i+k-1}dx\\ &= \int_0^1 \sum_{i=0}^p\binom{p}{i}(-1)^ix^{i+k-1}dx\\ &= \int_0^1 x^{k-1}(1-x)^pdx \end{split}$$ Then since $$\frac 1{(i+2)(i+4)}=\frac 1 2 \left(\frac 1 {i+2}-\frac 1 {i+4}\right)$$ we have $$a_p=\frac 1 2 (S_{2,p}-S_{4,p})=\frac 1 2 \int_0^1 (x-x^3)(1-x)^pdx$$ Thus,
$$\begin{split} \sum_{p=0}^n &a_p= \frac 1 2 \int_0^1(x-x^3)\sum_{p=0}^n(1-x)^pdx\\ &=\frac 1 2 \int_0^1(x-x^3)\frac{1-(1-x)^{n+1}}{x}dx\\ &=\frac 1 2 \int_0^1 (1-x^2)(1-(1-x)^{n+1})dx\\ \end{split}$$
Using Lebesgue's Dominated Convergence Theorem: $$\lim_{n\rightarrow+\infty}\sum_{p=0}^n a_p = \frac 1 2 \int_0^1 (1-x^2)dx=\frac 1 2 \left (1- \frac 1 3\right) = \frac 1 {3}$$