Yes, it is.  If $G$ is a discrete group acting on a simply-connected simplicial complex $X$, then a theorem of M. A. Armstrong says that there is a short exact sequence

$$1 \longrightarrow H \longrightarrow G \longrightarrow \pi_1(X/G) \longrightarrow 1,$$

where $H$ is the subgroup of $G$ generated by elements that fix a point in $X$.  The original reference is

M. A. Armstrong, On the fundamental group of an orbit space, Proc. Cambridge Philos. Soc. 61 (1965),
639–646.

See also my paper

A. Putman,
Obtaining presentations from group actions without making choices,
Algebr. Geom. Topol. 11 (2011), 1737-1766.

which proves a more specific result and contains a streamlined version of Armstrong's proof in Section 3.1 (which is really just a meditation on the usual facts about covering spaces, which the above restricts to if the action is free and thus $H = 1$).

In any case, if we apply this with $G = \text{Out}(F_n)$ and $X = X_n$ (Outer Space), then we see that $H = G$ since $G$ is generated by torsion elements, which fix points of Outer space.  The quotient is thus simply-connected, as desired.

A similar argument shows that the (ordinary, non-orbifold) fundamental group of the moduli space of curves is trivial.  This was originally proved (by this argument) in

Maclachlan, Colin, Modulus space is simply-connected, Proc. Amer. Math. Soc. 29 1971 85–86.