The question is a little hard to define because, as pointed out in the comments, it's not clear what the "outside volume" is, or how to define it in curved space to compare it to regular space. Still, I think there is a sense in which the answer is "yes".
A regular sphere has area $A = 4 \pi R^2$ and volume $V = 4 \pi R^3/3$, giving a relation $V = (1/6)\sqrt{A^3/\pi}$. We may ask whether there is some situation where a sphere would have a larger volume for a given area. If we imagine that someone standing on the surface of this sphere can only measure the area, they would infer the volume using the above formula, and they would be surprised to know that the interior volume is in fact larger.
There is a situation in which this can happen. For many years, cosmologists thought that the most likely shape for our universe was that of a 3-sphere, which at a fixed cosmological time has a metric given by
$$ds^2 = \frac{dr^2}{1-r^2} + r^2 (d\theta^2 + \sin^2 \theta d\varphi^2)$$
in suitable coordinates. The area of a sphere at a given radius is still $A = 4\pi R^2$, but following the standard methods of differential geometry the volume is
$$V = 4\pi \int_0^R dr\ \frac{r^2}{\sqrt{1-r^2}} = 2\pi \left(\arcsin R - R \sqrt{1-R^2}\right).$$
You can see by plotting that for any $R$, this volume is larger than the one given by the usual formula, even though the area is the same.
Edit in response to your edit: you ask whether it's possible to fit an elephant in a mouse sized box. Unless our conception of space radically changes (again), then the answer is clearly no. A mouse sized box is a box in which a mouse fits, i.e., its interior volume is that of a mouse. You're asking whether an object can have an interior volume larger than its interior volume; I hope it's clear that this is not possible. You can, however, fit an elephant in a box with the surface area of a mouse.