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, 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. 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.<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 maps from co-<a href="http://eom.springer.de/m/m064900.htm">Moore $M$-spaces</a> to $X$, which is, I suppose, a variation on your idea.<br>
<b>Question</b>: What is the co-Moore space for $\mathbb{Z}^n$?