There is a lot to unpack here, and probably this is a better question for math.SE. TL; DR version is: in exact arithmetic, this does not converge; see  Maxim Umansky's answer. In FP arithmetic, it will converge. I don't know to what, see the long version for an attempt.

Let's assume that it is possible to compute $\int_0^{\pi} f_n(x)\text{d}x$ exactly. I don't want to deal with the extra arithmetic coming from there, and it doesn't change the final answer much anyways.

After a while $\int_0^{\pi} f_n(x)\text{d}x$ will become small enough that $$fl(\sum_{n=1}^N \int_0^{\pi} f_n(x) \text{d}x+ \int_0^{\pi} f_{N+1}(x)\text{d}x)= fl(\sum_{n=1}^N \int_0^{\pi} f_n(x)\text{d}x)$$ where $fl$ is the rounding function. This is a common phenomenon in computational mathematics that you should be careful about.

(A well-known example is $\sum_{n=1}^{\infty} 1/n$ which diverges in exact arithmetic, but converges in FP arithmetic. If you reverse the ordering of the sum $\sum_{n=M}^{1} 1/n$ for $M\gg 1$, then you get the expected divergence result.)

Also, $f_n(x)$ itself can be thought as a fixed point iteration, and the integral $\int_0^{\pi} f_n(x)$ can be interpreted as testing the convergence of the fixed point iterations for all the initial guesses in the interval $[0,\pi]$ at the $n$-th iteration. The fixed point iteration $x=\sin(x)$ has the solution $x=0$ for any initial guess $x_0\in [0,\pi]$ (not too hard to prove, so I will skip). Hence, the integral $\int_0^{\pi} f_n(x)\text{d}x \to 0$ as $n\to\infty$. The upper bound for this integral is $\pi f_n(\pi/2)$.

Now, we can naively bound $\sin^n(\pi/2)$ ($\sin^n$ means $\sin$ composited with itself $n$ times) from below by $1/n$. Here is a proof by induction: 

$n=1$ case: $\sin^1(\pi/2) \geq 1/1$. Hence, we are good here.

Inductive hypothesis: Assume for $n=k$, $\sin^n(\pi/2)\geq 1/n$.

$n=k+1$ case: Consider $\begin{align}\sin(\sin^k(\pi/2)) - 1/(k+1) &= \sin(\sin^k(\pi/2)) - 1/k + 1/(k^2+k) \\
&\text{by the inductive hypothesis and since sine is an increasing func. in } [0,1]\\
&\geq \sin(1/k) - 1/k + 1/(k^2+k) \\ &\text{by Taylor's remainder theorem}\\
&\geq 1/k -1/(6k^3) -1/k +1/(k^2+k)\\  &\text{for any } k\geq 1\\
&\geq 0.
\end{align}$

So, the lower bound for the sum $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ is $\pi\sum_{n=1}^{\infty} 1/n$, which will definitely not increase after $n=4.5\times 10^{15}\approx \varepsilon_{\text{mach}}$ since $fl(1+\varepsilon_{\text{mach}}) = fl(1)$. I would expect $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ to "converge" earlier than that (the word "converge" is in quotation marks, because it is not a real convergence). 

On the other hand, we can use $1$ as an upper bound to the upper bound $f_n(\pi/2)$ (since the fixed point iterations converge to zero). Then we see that $$4.5\pi\times 10^{15}\approx\pi\sum_{n=1}^{\infty} 1 \geq \pi\sum_{n=1}^{\infty} f_n(\pi/2) \geq \pi\sum_{n=1}^{\infty} 1/n \approx \pi(\ln(4.5\times 10^{15}) + \gamma),$$
where $\gamma$ is the Euler–Mascheroni constant.

Hence, the sum $\sum_{n=1}^{\infty} \int_0^{\pi} f_n(x)$ definitely has a finite limit when calculated on a computer using floating point arithmetic and that value is no larger than $4.5\pi\times 10^{15}$. I would expect it to be slightly larger $\pi\ln(4.5\times 10^{15})$, but I cannot prove that rigorously. That is more of an educated guess.