This might not be the final answer, but it might help you understand the issue. I will rely on other users to help me get the whole picture.
There is a certain rule of thumb as to know when a quantity is a scalar or not: if the indices go through all space-time dimensions, i.e., $\alpha=0,1,2,3$, then the expression is probably a scalar. Your formula is not a scalar because $i=1,2,3$, so when performing a change of reference system, the zeroeth component of $A$ and $\partial$ will mess up your formula: it changes in a nasty way.
On the other hand, something like $\partial_\alpha A^\alpha$, where $\alpha=0,1,2,3$ is in general a scalar, because the index can take the four values. The thing is: lets say you write, in classical mechanics, an equation like $ma_x=F_x$, $ma_y=F_y$, $ma_z=2F_z$. This is clearly a "bad" equation, because it's not symmetric: the $z$ component of your vectors is trated very differently compared to the others. If you and me use this formula in order to get physical predictions, and we both use different systems of reference, then we will get very different predictions. This formula can't work in general. Let's say, we by chance pick systems of reference such that my $y$ axis is your $z$ axis, and the $x$ coincides; also imagine that in a particular experiment there is only a force in the $x$ direction; in this case our predictions will agree. This is similar to what happened in physics: we had equations that were not invariant, but our systems of reference were so similar that we couldn't detect any flaw in them. If you take a very fast-moving system of reference, then you will easly see that newtonian mechanics predictions disagree between observers.
In SP, the time component of vectors is just as important as the others, so your formulas should be symmetric w.r.t the four components. If you had in Newtonian mechanics a formula that tells you something about the $x$ and the $y$ component of some vector, then there should be another analogous formula for the $z$ component; otherwise a rotation would make your formula nonsense.
In this sense, you should be able to tell if this formluas are "good" formulas or not:
1) $\partial_\alpha A^\alpha=A^0$
2) $\partial_\alpha\partial^\alpha T^\beta=j^\beta$
3) $j^0=|\phi|^2\ \&\ j^i=\phi^*\partial^i\phi-\phi\partial^i\phi^*$
4) $p_0^2=p_i p^i-p_0 p_x p_y p_z$
and so on. Remember that a greek index is will always take four values and a latin one only three.
So remember: invariant formulas are symmetric w.r.t. the four components of vectors, and everything should be writen with greek indices, not latin ones. If a formula includes latin indices but you don't see the $0$ component around, this will mean that the formula is not invariant. This might be tricky, because you will sometimes find things like $\partial_t \rho+\nabla\cdot \vec j=0$, and this might look non-invariant, but it really is, provided you have $\rho=j^0$.
I don't know why you make any distiction between a tensor and a Lorentz-tensor: they are the same thing. While studying mathematics, we talk about general tensors because we don't care about physical theories, but when the vector space of interest is space-time, then the tensor is a Lorentz-tensor. In general we never say "Lorentz-tensor", but simply "tensor".
PS: only 2) is invariant. Can you tell why the others are not?