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Quite apart from the fact that, as Charles Rezk pointed out, you wouldn't get a group, your problem is that $T^n$ is not in general a co-Moore $M$-spaces. Therefore  Eckmann-Hilton duality breaks down, and there would be no (co)homology theory dual to such a "homotopy theory". Thus, a theory of homotopy classes of pointed maps from $T^n$ to $X$ would be much less interesting than a theory of homotopy classes of pointed maps from $S^n$ to $X$.

Let's think about what else 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 $\mathbb{Z}$. What makes the ring of integers special? That it is an initial object for the category of rings. One consequence is that (in principle at least) you can define homotopy with arbitrary coefficients by extension of scalars. Rational homotopy theory is constructed this way.
What would go wrong if you tried the same thing for $T^n$? 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 would be no rational theory for example. The theory would be less fundamental and less well-behaved; therefore less worth investigating.
On the other hand, the study of homotopy classes of pointed maps from a co-Moore space other than $S^n$ to $X$ does lead to the useful theory of homotopy with coefficients. I believe these classify $X$ up to homotopy equivalence.

Your problem is that $T^n$ is not in general a co-Moore space. Therefore Eckmann-Hilton duality breaks down, as the dual spaces no longer form a spectrum, and there would be no (co)homology theory dual to such a "homotopy theory". Thus, a theory of homotopy classes of pointed maps from $T^n$ to $X$ would be much less interesting than a theory of homotopy classes of pointed maps from $S^n$ to $X$.

On the other hand, the study of homotopy classes of pointed maps from a co-Moore space other than $S^n$ to $X$ does lead to useful theories of homotopy with coefficients. I believe these classify $X$ up to homotopy equivalence.

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 here. 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.
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.
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-Moore $M$-spaces to $X$, which is, I suppose, what you really wanted.

Quite apart from the fact that, as Charles Rezk pointed out, you wouldn't get a group, your problem is that $T^n$ is not in general a co-Moore $M$-spaces. Therefore Eckmann-Hilton duality breaks down, and there would be no (co)homology theory dual to such a "homotopy theory". Thus, a theory of homotopy classes of pointed maps from $T^n$ to $X$ would be much less interesting than a theory of homotopy classes of pointed maps from $S^n$ to $X$.

Let's think about what else 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 $\mathbb{Z}$. What makes the ring of integers special? That it is an initial object for the category of rings. One consequence is that (in principle at least) you can define homotopy with arbitrary coefficients by extension of scalars. Rational homotopy theory is constructed this way.
What would go wrong if you tried the same thing for $T^n$? 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 would be no rational theory for example. The theory would be less fundamental and less well-behaved; therefore less worth investigating. 
On the other hand, the study of homotopy classes of pointed maps from a co-Moore space other than $S^n$ to $X$ does lead to the useful theory of homotopy with coefficients. I believe these classify $X$ up to homotopy equivalence.

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 here. 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.
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.
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-Moore $M$-spaces to $X$, which is, I suppose, a variation on your idea.
Question: What is the co-Moore space for $\mathbb{Z}^n$?

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 here. 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.
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.
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-Moore $M$-spaces to $X$, which is, I suppose, what you really wanted.