For $0<x<\pi/2$ there is $\sin x<x<\tan x$, so for these $x$, an upper bound is
while for $-\pi/2<x<0$ there is $\tan x<x<\sin x$, so for these $x$ an upper bound is
It looks like there is not a "simple" function which does the job on both sides of $0$ using the same formula. And also $\arcsin(\tan(x))$ is more complicated than the function you're trying to bound. Still, the taylor series for it may give a polynomial upper bound for $x>0$ [I haven't checked this].
EDIT: I just found some terms of the taylor series of $\arcsin(\tan(x))$. It starts out
This will be greater than $\arcsin(x)$ if we use all the terms; if one wants the best upper bound from this, clearly one should not keep adding positive terms, but stop as soon as the series "beats" the series for $\arcsin(x)$. I haven't checked, but it looks like all we get from the taylor series for $\arcsin(\tan(x))$ is the simple statement
which seems to hold for reasonably small positive $x$.
I guess what one would really want is a sequence of upper bounds that converged down to $\arcsin x$ for positive $x$. I don't know how to construct that.