If your triangulated category arises a homotopy category of a stable oo-category (as most well-known examples do), and the commutative chain of squares $A_i \to B_i$ arises from a diagram in the oo-category (as many natural diagrams do), the answer is yes. An exact triangle is given by a hocolim of the diagram $0 \leftarrow A_i \to B_i$, and hocolims commute past each other, so

$$ hocolim ( 0 \leftarrow hocolim_{\mathbb{Z}_{\geq 0}} A_i \to hocolim_{\mathbb{Z}_{\geq 0}} B_i) \simeq hocolim_{\mathbb{Z}_{\geq 0}} hocolim ( 0 \leftarrow A_i \to B_i) \simeq hocolim_{\mathbb{Z}_{\geq 0}} C_i $$

i.e., we have an exact triangle

$$ hocolim_{\mathbb{Z}_{\geq 0}} A_i \to hocolim_{\mathbb{Z}_{\geq 0}} B_i \to hocolim_{\mathbb{Z}_{\geq 0}} C_i $$

If your triangulated category arises a homotopy category of a stable $\infty$-category (as most well-known examples do), and the commutative chain of squares $A_i \to B_i$ arises from a diagram in the $\infty$-category (as many natural diagrams do), the answer is yes. An exact triangle is given by a hocolim of the diagram $0 \leftarrow A_i \to B_i$, and hocolims commute past each other, so

$$ hocolim ( 0 \leftarrow hocolim_{\mathbb{Z}_{\geq 0}} A_i \to hocolim_{\mathbb{Z}_{\geq 0}} B_i) \simeq hocolim_{\mathbb{Z}_{\geq 0}} hocolim ( 0 \leftarrow A_i \to B_i) \simeq hocolim_{\mathbb{Z}_{\geq 0}} C_i $$

i.e., we have an exact triangle

$$ hocolim_{\mathbb{Z}_{\geq 0}} A_i \to hocolim_{\mathbb{Z}_{\geq 0}} B_i \to hocolim_{\mathbb{Z}_{\geq 0}} C_i $$