The growth rate is exponential. Specifically, I will prove the bound $J_n \leq \frac{(n+1)^{n} 2^n}{(n+1)!}$, which grows like $(2e)^n$. I don't expectcan also show that this is$J_n \geq 2^n$. I would guess that neither of these are the true rate of growth.
I'll also give an explicit recursion for the $J_n$, which makes it clear that they are all in $\mathbb{Q}[\pi]$ and should make it easy to compute them recursively.
It is convenient to define $J_n(b)$ to be $J_n$ with every upper bound $1$ replaced by $b$. For example, $$J_2(b) = \int_{t_2=0}^b d t_2 \int_{t_3=0}^{t_2} d t_3 \frac{1}{\sqrt{b-t_2}} \left( \frac{1}{\sqrt{b-t_3}} + \frac{1}{\sqrt{t_2-t_3}} \right)$$ By homogeneity, $J_n(b) = J_n \cdot b^{n/2}$. We set $J_0(b)=1$.
Here is the recursion in the form which is easiest to prove: $$J_n(b) = \sum_{k\geq 1} \frac{1}{k!} \sum_{\begin{matrix} n_1+n_2 + \cdots + n_k = n \\ n_1, n_2, \ldots, n_k >0 \end{matrix}} \prod_{i=1}^k \int_{c_i=0}^b \frac{J_{n_1-1}(c_i) d c_i}{\sqrt{b-c_i}}$$
Sketch of proof of recursion: The left hand side is a sum of $n!$ terms which are naturally indexed by functions $f: \{ 2,3,\ldots, n+1 \} \to \{1,2,\ldots, n \}$ obeying $f(i) \lt i$. The term corresponding to $f$ has integrand $1/\prod \sqrt{t_{f(i)} - t_i}$, where $t_1=b$.
Group together those summands for which $f^{-1}(1)$ has size $k$; let $(u_1, u_2, \ldots, u_k) = f^{-1}(1)$. For any $x \in \{2,3, \ldots, n \}$, there is precisely one $u_i$ such that $u_i \in \{ x, f(x), f(f(x)), f^3(x), \ldots \}$ Say that "$x$ passes through $u_i$". Let $n_i$ be the number of $x$ which pass through $u_i$. We group together all terms on the left with the same unordered list of multiplicities $(n_1, n_2, \ldots, n_k)$. On the right, group together all terms which differ only by permuting $(n_1, n_2, \ldots, n_k)$. I claim that the corresponding sums on the two sides match up.
An example is probably clearer than a proof here. Take $n=3$ and look at the partition $3=2+1$. On the left hand side, we have terms accounting for $(f(2), f(3), f(4))$ equal to: $(1, 1, 3)$, $(1,1,2)$ and $(1,2,1)$. On the right hand side, we have $2$ copies of $$\frac{1}{2!} \int_{c_1=0}^b \int_{c_2=0}^b \int_{t=0}^{c_1} \frac{d c_1 dc_2 dt}{\sqrt{b-c_1} \sqrt{b-c_2} \sqrt{c_1-t}}$$ The two copies cancel the $1/2!$. We are integrating over the region $0 \leq t \leq c_1$, $0 \leq c_2$. There are three possible orderings for the integrands: $0 \leq c_2 \leq t \leq c_1$, $0 \leq t \leq c_2 \leq c_1$ and $0 \leq t \leq c_1 \leq c_2$, and these correspond to the three functions $f$ above.
The confusing thing in writing out a general proof is noticing that the $1/k!$ and the number of orderings of the partition $n_1+n_2+ \cdots + n_k$ exactly cancel to eliminate any double counting. Since you tagged this with "feynmann-integrals", I assume you are used to the way these sort of symmetry factors work. $\square$
Simplifying the recursion We first plug in $J_m(b) = J_m \cdot b^{m/2}$ to obtain $$J_n = \sum_{k\geq 1} \frac{1}{k!} \sum_{\begin{matrix} n_1+n_2 + \cdots + n_k = n \\ n_1, n_2, \ldots, n_k >0 \end{matrix}} \prod_{i=1}^k J_{n_1-1} \int_{c_i=0}^1 \frac{ d c_i \ c_i^{(n_i-1)/2}}{\sqrt{b-c_i}}$$
The integrals are now simple one dimensional integrals and can be evaluated: $$J_n = \sum_{k\geq 1} \frac{1}{k!} \sum_{\begin{matrix} n_1+n_2 + \cdots + n_k = n \\ n_1, n_2, \ldots, n_k >0 \end{matrix}} \prod_{i=1}^k \left( J_{n_1-1} \cdot \frac{\sqrt{\pi}\ \Gamma((n_i+1)/2)}{\Gamma((n_i+2)/2)} \right)$$ Note that $\left( \frac{\sqrt{\pi}\ \Gamma((n_i+1)/2)}{\Gamma((n_i+2)/2)} \right)$ is rational if $n_i$ is odd and is a rational multiple of $\pi$ if $n_i$ is even, so this shows that all your polynomials are polynomials in $\pi$. (I am feeling nicely superior to Mathematica now. )
AsymptoticUpper bounds We have $ \frac{\sqrt{\pi}\ \Gamma((n_i+1)/2)}{\Gamma((n_i+2)/2)} \leq 2$. So $J_n$ is bounded by $K_n$, where $K_n$ is defined by the recursion $$K_n = \sum_{k\geq 1} \frac{1}{k!} \sum_{\begin{matrix} n_1+n_2 + \cdots + n_k = n \\ n_1, n_2, \ldots, n_k >0 \end{matrix}} \prod_{i=1}^k \left( 2 \cdot K_{n_1-1} \right)$$ Define the generating function $g(x) = \sum_{n=0}^{\infty} K_n x^n$. Then this shows $$g(x) = \exp(2 x g(x))$$
We thus see that $g(x) = -W(-2x)/(2x)$ where $W$ is the Lambert $W$ function and so $$K_n = \frac{(n+1)^{n} 2^n}{(n+1)!}$$ as promised.
Lower bound We can use a similar trick to get a lower bound. Note that $\frac{\sqrt{\pi} \Gamma((n+1)/2)}{\Gamma((n+2)/2)} \geq \frac{2}{n}$. So $J_n \geq L_n$ where $L$ is defined recursively by $L_0=1$ and $$L_n = \sum_{k\geq 1} \frac{1}{k!} \sum_{\begin{matrix} n_1+n_2 + \cdots + n_k = n \\ n_1, n_2, \ldots, n_k >0 \end{matrix}} \prod_{i=1}^k \left( \frac{2}{n} L_{n_1-1} \right)$$
Putting $L(x) = \sum_{n \geq 0} L_n x^n$, we see that $$L(x) = \exp\left( 2 \int_{t=0}^x L(t) dt \right).$$ This integral equation has the unique solution $L(x) = 1/(1-2x)$, so $L_n = 2^n$, as promised.