"Distance" in this case means the metric function, not "ordinary distance". Let's take a point "as far away as possible" from the origin, say (1,1,1).

Then d((0,0,0),(1,1,1)) = max{|0 -1|,|0-1|,|0-1|} = max{|-1|,|-1|,|-1|} = max{1,1,1} = 1.

Convince yourself that if (x,y,z) is any point with two of the 3 coordinates with absolute value less than 1, and the third coordinate with absolute value equal to 1, that d((0,0,0),(x,y,z)) = 1, as well.

Hopefully this allows you to see that the boundary of N(0,1) is precisely the faces of said cube, so all the interior points comprise N(0,1).