I haven't looked at the paper, but I can help calculate $[\mathbb T^3, U]$ - which is $\mathbb Z^{\oplus 4}$. Perhaps you wanted $[\mathbb T^3, SU] = \mathbb Z$.
I will first give an elementary argument, then a more algebraic one that uses Bott periodicity.
Argument 1. First consider the special unitary group. As is discussed at MO there is a cell structure for $SU(n)$ having 4-skeleton $SU(2) = S^3$. Every map $\mathbb T^3 \to S^3$ is homotopic to a cellular map, which thus factors through $\mathbb T^3 /(2-\text{skeleton}) = S^3$. Since there are no 4-cells we actually have $$\mathbb Z = \pi_3(S^3) = [\mathbb T^3, SU].$$ This reasoning is pretty common, it's similar to, say, MSE. As for $U(n)$, one has $U(n) \simeq SU(n)\times S^1$ so $$[\mathbb T^3, U] = [\mathbb T^3, SU] \times H^1(\mathbb T^3;\mathbb Z) = [\mathbb T^3, SU] \times \mathbb Z^3.$$
We have just used that $[X, S^1] = H^1(X,\mathbb Z)$, see MSE.
Here is a more algebraic argument.
Argument 2. From MSE we have $\Sigma(X\times Y) \simeq \Sigma X \vee \Sigma Y \vee \Sigma(X\wedge Y)$
Thus for $\mathbb T^3 = S^1\times \mathbb T^2$ we calculate
$$\Sigma(\mathbb T^3) \simeq S^2\vee\Sigma \mathbb T^2 \vee\Sigma^2\mathbb T^2$$
It is a common exercise that $\Sigma \mathbb T^2 \simeq S^2\vee S^2 \vee S^3$ (this must be on MSE somewhere...) thus
$$\Sigma\mathbb T^3 \simeq S^2 \vee (S^2\vee S^2 \vee S^3)\vee(S^3\vee S^3\vee S^4) = (S^2)^{\vee 3}\vee(S^3)^{\vee 3}\vee S^4$$
Now we can use Bott perioidicty again $$[\mathbb T^3, U] = [\mathbb T^3, \Omega^2U] = [\Sigma^2 \mathbb T^3, U].$$
Thus $$[\mathbb T^3, U] = \pi_3(U)^{\oplus 3} \oplus \pi_4(U)^{\oplus 3} \oplus \pi_5(U) = \mathbb Z^{\oplus 4}.$$
