Is the spacetime interval a tensor?

Edit: I no longer agree with my answer, and I think it would be appropriate to accept one of the other answers. Depending on what you mean by the spacetime interval, it's either a non-local, non-tensorial thing (if we're talking about $\Delta s$), or a 2-form that happens to be preserved by Lorentz transformations (if we're talking about $ds^2$).

The spacetime interval is Lorentz invariant. That is, it is a scalar. A scalar is a particular case of a tensor. One could call it a (0,0)-tensor (as in zero covariant and zero contravariant indices).

The answer is technically yes, as has been lined out, but in regards to the second part of the question the answer is an emphatic NO. It should never be treated as a tensor because that only works in flat space and will only lead to future misunderstandings and lots of wasted time.

To give a longer answer: The great thing about most tensors in special relativity is that they turn into similar tensors in general relativity because they are local and locally space-time still looks like Minkowski space. The reason this does not work with the space-time interval is that it is actually non-local. Here local means that it should only depend on what happens in a small neighborhood around the single point you are considering. If you consider the space-time distance between two points it kind of depends on two different points as well as on everything that happens in between.

What you should think about the space-time interval instead is as the length of a curve taken in a specific invariant way. It depends on the curve taken and thus on the end-points and everything inbetween. It just so happens in flat Minkowski-space that for any pair of points there is an obvious curve between two points in the form of a straight line.

The spacetime interval is a bilinear map that takes two (relative position) 4-vectors and produces a scalar. This means that is a rank 2 tensor (more specifically type $(0,2)$). In most basic special relativity context, the two 4-vectors are taken to be two copies of the same position 4-vector, but this does not change the mathematical nature of the underlying object.