Let $$f_n(x,y,z)= \frac{\arctan^a (|x|+|y|)e^z}{n^4+4x^2+y^2} $$ on $$D=\{(x,y,z):1<x^2+y^2+z<2\}$$
I want to determine for which $a\in \mathbb{R}$ we have $f_n\in L^1(D)$ for all $n\geq 1.$
The following is in part my attempt at a solution and in part a long comment to the solution proposed below.
The first thing that comes to mind is that we may rewrite $D$ splitting it in the disjoint union of subsets $$D_{z=0}\cup D_{z>0}\cup D_{z<0}$$
Now, for $z=0$, we have to integrate $$\frac{\arctan^a (|x|+|y|)}{n^4+4x^2+y^2}$$ on $$\{0<x^2+y^2<2\}$$ which is the region delimited by a triagle with sides $\sqrt{2}$, thus a bounded set in $\mathbb{R}^2;$ since the function is continous on this triangle we obtain that is integrable for all $a.$
For $z>0,$ we have $$D_{z>0}=\{(x,y,z): \ 0<x^2+y^2<2 \ , \ z>0 \ , \ 1-(x^2+y^2)<z<2-(x^2+y^2) \ \}$$ But then we see that $$D_{z>0}\subset [0,\sqrt{2}] \times [0,\sqrt{2}] \times [0,2]$$ and thanks to the $n^4$ factor the function is continuous on this region, so that it is integrable on it for all $a.$
It remains to treat the case $z<0.$
Since $1-(x^2+y^2)<z<0,$ we must have $(x^2+y^2)>1.$ So $D_{z<0}$ is given by the unbounded region of space $$D_{z<0}=\{(x,y,z): x^2+y^2>1 , \ z<0, \ 1-(x^2+y^2)<z<2-(x^2+y^2) \}$$ (which is an open set of $\mathbb{R}^3$ and thus measurable.)
I also notice that $|f_n(x,y,z)|=f_n(x,y,z).$
Moreover, $$f_n(-x,-y,z)=f_n(x,y,z)$$ for all $z$ and since $(\pm x, \pm y, z)\in D_{z<0} \implies (\mp x, \mp y, z)\in D_{z<0},$ this should imply that $$\int_{D_{z<0}}f_n=4\int_{E} f_n$$ where $$E={D_{x>0,y>0,z<0}}$$
Now we can pass to cylindrical coordinates and the domain of integration becomes:
$$F=\{(\theta,\rho,z): \theta \in[0,\pi/2], \ \rho \in (1,+\infty), \ z \in (1-\rho^2,2-\rho^2) \}$$
and we have to evaluate:
$$\int_0^{\pi/2}\int_1^{+\infty} \frac{\arctan^a (\rho(|cos\theta|+|sin\theta|))}{n^4+4\rho^2\cos^2\theta+\rho^2\sin^2\theta} \rho d\rho d\theta \int_{1-\rho^2}^{2-\rho^2} e^{-|z|} dz$$
This one seems ripe for cylindrical coordinates $r,\theta,z.$ So for fixed $r,\theta,$ we see $1-r^2<z<2-r^2.$ The $z$ integral thus equals $e^{2-r^2}-e^{1-r^2}.$ Your integral becomes
$$\int_0^{2\pi}\int_0^\infty \frac{\arctan^a (r|\cos\theta|+r|\sin \theta|)}{n^4+4r^2\cos^2 \theta +r^2\sin^2 \theta}(e^{2-r^2}-e^{1-r^2})r\,dr\,d\theta.$$
Now
$$r \le r|\cos\theta|+r|\sin \theta| \le \sqrt 2 r.$$
Therefore the $\arctan$ business lies between $\arctan^a(r)$ and $\arctan^a(\sqrt 2 r).$ In the denominator, note
$$1 \le n^4+4r^2\cos^2 \theta +r^2\sin^2 \theta \le n^4 +4r^2.$$
Using these estimates gets rid of the $\theta$ integral altogether. As for the $r$ integral, let's break it into $0<r<1$ and $r\ge 1.$ When is
$$\int_0^1 \frac{\arctan^a(\sqrt 2 r)}{1}(e^{2-r^2}-e^{1-r^2})r\,dr <\infty?$$
Now $\arctan(\sqrt 2 r)\le Cr$ for some positive constant $C.$ So when is
$$\int_0^1 \frac{C^ar^a}{1}(e^{2-r^2}-e^{1-r^2})r\,dr <\infty?$$
The exponential stuff does not affect convergence here. So we're left wondering about $\int_0^1 r^{a+1}\,dr.$ That converges for $a>-2.$ The logic then says $a>-2$ suffices for convergence of the $0<r<1$ integral. But it's also necesary; verify that we didn't give up much in the "sufficient direction".
As for the range $1<r<\infty,$ this is the easy part. It always converges. The $\arctan $ stuff is simply bounded here, so forget about it. This integral will converge if
$$\int_0^\infty (e^{2-r^2}-e^{1-r^2})r\,dr $$
converges. Which it does, no contest.
Final answer: The original integral converges iff $a>-2.$