Let's think about what makes $S^n$ special, and why homotopy classes of maps from $S^n$ to $X$ are more likely to provide a firm basis for a robust mathematical theory then maps from $T^n$ to $X$. The key property of $S^n$, as far as homotopy is concerned, is that it has a single nonvanishing cohomology group, which is $\mathbb{Z}$. Read why I say this <a href="http://books.google.com/books?id=XxbW1jzB1qwC&pg=PA460&lpg=PA460&dq=Eckmann-Hilton+duality&source=bl&ots=OePoDequhs&sig=H5URNjMpXwHtvPsUV6Jd9xRU8rQ&hl=en&ei=IOuDTOW2IKeynAeck4jeAQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CD8Q6AEwCQ#v=onepage&q=Eckmann-Hilton%20duality&f=false">here</a>. What makes the ring of integers special? That it is an initial object for the category of rings. So you can define homotopy with arbitrary coefficients by extension of scalars, for instance. I believe these classify $X$ up to homotopy equivalence.<br>
What would go wrong if you tried the same thing for $T^n$? First, as Charles Rezk pointed out, you wouldn't get a group. But beyond that, the sole nonvanishing cohomology class $\mathbb{Z}^n$ of $T^n$ is not in general an initial object for the category of rings (even if we forget that the "$n$" in $\mathbb{Z}^n$ changes as the "$n$" in $T^n$ changes, which is a total disaster). So I don't think you would have any hope for a sensible theory with arbitrary coefficients, even in principle. The theory would be less fundamental and less well-behaved; therefore less worth investigating. Via Eckmann-Hilton duality, what I have just told you is why I think it is more fundamental to consider cohomology with integral coefficients as a basis for a theory than to consider cohomology with coefficients in some other ring as a basis for a theory.<br>
On the other hand, naive extension of scalars isn't typically a good way to get homotopy with coefficients (although it works fine for rational coefficients). One instead considers homotopy classes of maps from co-<a href="http://eom.springer.de/m/m064900.htm">Moore $M$-spaces</a> to $X$, which is, I suppose, what you really wanted.