I believe you meant to write $\mathbf{T}_g:=\prod_{i=1}^n T_{g(i)}$ so maybe a better symbol would be $\mathbf{T}_g^n:=\prod_{i=1}^n T_{g(i)}$.

Now, follwing @DongryulKim's suggestion, we can see why your claim holds by observing how $\mathbf{T}^n$ is formed.

Let us characterize functions $g \in G(n,d)$ by the values they take $g_{i_1, i_2, \dots , i_d} \in G(n,d)$, meaning $g_{i_1, i_2, \dots , i_d}(k) = i_k$ with $k \in {1, 2, \dots , n}$ and $i_k \in {1,2,\dots,d}$.

Clearly, when $N=1$ the components run over all the functions in $G(1,d) = \{g_1, g_2 , \dots, g_d \}$ since

$$\mathbf{T}^1 = (T_1, T_2, \dots , T_d) = \left(T_{g_1(1)}, T_{g_2(1)}, \dots, T_{g_d(1)} \right ) = \left(\mathbf{T}_{g_1}^1, \mathbf{T}_{g_2}^1, \dots, \mathbf{T}_{g_d}^1 \right ) $$

Let us assume now that this also holds for $N=n$, that is
$$\mathbf{T}^n = (\mathbf{T}_{g_{1,1,\dots,1}}^n, \dots, \mathbf{T}_{g_{i_1,i_2,\dots,i_n}}^n, \dots , \mathbf{T}_{g_{d,d,\dots,d}}^n)$$

Now, going to $N=n+1$ we have

$$\mathbf{T}^{n+1} = (T_1 \mathbf{T}_{g_{1,1,\dots,1}}^n, \dots ,T_1 \mathbf{T}_{g_{d,d,\dots,d}}^n ,
T_2 \mathbf{T}_{g_{1,1,\dots,1}}^n, \dots, T_2 \mathbf{T}_{g_{d,d,\dots,d}}^n , \cdots ,
T_d \mathbf{T}_{g_{1,1,\dots,1}}^n, \dots, T_d \mathbf{T}_{g_{d,d,\dots,d}}^n
)$$

It is easy now to see from this arrangement that the first block of components corresponds to terms $\mathbf{T}_g^{n+1}$ for functions $g \in G(n+1,d)$ with $g(1) = 1$ and $g(i) = g'(i-1)$ for any function $g' \in G(n,d)$ and $i>1$. Similarly, the second block corresponds to terms $\mathbf{T}_g^{n+1}$ for functions $g \in G(n+1,d)$ with $g(1) = 2$ and $g(i) = g'(i-1)$ for any function $g' \in G(n,d)$ and $i>1$. Following this all the way to the last block, where $g(1) = d$ we see that we have covered all the functions in G(n+1,d). Thus

$$\mathbf{T}^{n+1} = (\mathbf{T}_{g_{1,1,\dots,1}}^{n+1}, \dots,\mathbf{T}_{g_{i_1,i_2,\dots,i_d}}^{n+1},\dots,\mathbf{T}_{g_{d,d,\dots,d}}^{n+1}
)$$

So, in general, we have

$$\mathbf{T}^n = (\mathbf{T}_{g_{1,1,\dots,1}}^n, \dots ,\mathbf{T}_{g_{i_1,i_2,\dots,i_d}}^n,\dots,\mathbf{T}_{g_{d,d,\dots,d}}^n
)$$

from which we can get

$$
\|\mathbf{T}^n\|^2=\sum_{(i_1,...,i_n) \in \{1,...,d\}^n}\|\mathbf{T}_{g_{i_1,...,i_n}}^n\|^2 =
\sum_{g\in \mathbf{G}(n,d)}\|\mathbf{T}_g^n\|^2
$$