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

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