The “boundary” of the Universe only exists in particular spacetimes, and probably not in ours. In flat spacetime, you can get as distant from any point you want and still never encounter a boundary. In some other curved spacetimes, like with simple global curvature that’s constant-positive or constant-negative, you can end up turning around and not reaching your destination, or reaching your origin again after traveling far enough. Even in more exotic spacetimes (like a toroid, where one of the dimensions is periodic in a particular way), there aren’t boundaries - you might end up back at your origin again, or never see your origin again, but in most cosmologically-valid spacetimes you never hit a hard-stop boundary.
Contrary to popular belief (I’m not going to bother getting a source; as a relativist I get asked plenty of questions from laypeople on the order of “what’s space like outside the Universe”), there isn’t space outside the Universe. Our space is roughly flat, so it extends ad infinitum in all directions, and you can reach any point in space by pointing yourself towards it and flying (avoiding things like planets and stars on the way). For there to be a place that is “outside the Universe”, the Universe would have to be bounded in a way that makes it so that some events in $\mathbb{R}^{1,3}$ (or whatever you want to call the set of events in spacetime) not actually be valid vectors that represent points that “exist”.
As far as we know, such points don’t exist at a cosmological scale. But the closest we can get is points that, in an external coordinate system, are inside a black hole. In spherical coordinates the Schwarzschild metric is
$$\mathrm{d}s^2=k_Sc^2 \mathrm{d}t^2-\frac{\mathrm{d}r^2}{k_S}-r^2 \mathrm{d}\theta^2-r^2\sin^2 \mathrm{d}\theta\phi^2.$$
Here, $k_S$ is short hand for $1-\frac{2GM}{rc^2}=1-\frac{r_s}{r}=1-\frac{v_e^2}{c^2}$, where $r_s$ is the Schwarzschild radius (event horizon radius) of the black hole. Even without very advanced calculus you can tell that this metric behaves badly at $r=r_s$: you end up dividing by zero. That’s a bit of an issue. That surface is the event horizon, which causally disconnects the interior of the black hole from the exterior - observers can see themselves cross the event horizon in a finite amount of proper time, but external spectators will see it take an infinite amount of coordinate time.
The resulting edge of the Universe (if we’re to interpret the Universe as being bounded as the exterior of the sum of all black holes) is that time dilation increases to infinity and spatial contraction - distances appearing to get shorter and shorter on the part of the observer falling in - also goes to infinity. Technically this does result in the loss of the time dimension at the event horizon surface, so you weren’t entirely wrong to think that spacetime gets degenerated into lower dimensions at the boundary. It’s a bit more complicated than the contraction of 3-space into 0-space, but the boundary we’ve defined is as close as we can get to the one you’re thinking about anyways.
