5. Connecting the centers of the faces of a regular polyhedron, we get a new regular polyhedron, which we call the positive sub-polyhedron of the original polyhedron. A cube $T_{1}$ has a surface area of $a_{1}=16$, its positive sub-polyhedron is $T_{2}$, with a surface area of $a_{2}$, the positive sub-polyhedron of $T_{2}$ is $T_{3}$, with a surface area of $a_{3}, \cdots \cdots$ and so on. Let the surface area of the $n$-th positive sub-polyhedron be $a_{n}$. Then
$$
\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=
$$
5. $18+3 \sqrt{3}$.

Given that $T_{1}, T_{3}, \cdots$ are cubes, and $T_{2}, T_{4}, \cdots$ are regular octahedra. Let the side length of $T_{i}$ be $b_{i}$. From Figure 4, it is easy to see that $b_{2}=\frac{\sqrt{2}}{2} b_{1}$. From Figure 5, we calculate that $C H=\frac{1}{2} A C=\frac{\sqrt{2}}{2} b_{2}$,
$$
b_{3}=M N=\frac{2}{3} C H=\frac{\sqrt{2}}{3} b_{2} \text {. }
$$

It is easy to see that for any natural number $n$, we have
$$
b_{2 n}=\frac{\sqrt{2}}{2} b_{2 n-1}, b_{2 n+1}=\frac{\sqrt{2}}{3} b_{2 n} \text {. }
$$

Given $a_{1}=6 b_{1}^{2}, a_{2}=8 \times \frac{\sqrt{3}}{4} b_{2}^{2}=\sqrt{3} b_{1}^{2}, a_{2}=\frac{\sqrt{3}}{6} a_{1}$,
$$
a_{3}=6 \times \frac{2}{9} b_{2}^{2}=\frac{4}{3} b_{2}^{2}=\frac{2 \sqrt{3}}{9} a_{2} \text {. }
$$

Similarly, $a_{2 n}=\frac{\sqrt{3}}{6} a_{2 n-1}, a_{2 n+1}=\frac{2 \sqrt{3}}{9} a_{2 n}$.
Thus, we have $\frac{a_{2 n+1}}{a_{2 n-1}}=\frac{1}{9}, \frac{a_{2 n+2}}{a_{2 n}}=\frac{1}{9}$.
Therefore, $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)$
$$
=\frac{a_{1}}{1-\frac{1}{9}}+\frac{a_{2}}{1-\frac{1}{9}}=18+3 \sqrt{3} .
$$