In the lattice $\mathbb{Z}^d$, consider the norm $\vert\vert \cdot \vert \vert \colon \mathbb{Z}^d \rightarrow [0,+\infty)$, given by $$\vert \vert \lambda \vert\vert:= \displaystyle\max_{1 \leq i \leq d} |\lambda_i|,$$ where $\lambda = (\lambda_1, \dots, \lambda_d)$.
We say that $\lambda, \lambda^{\prime} \in \mathbb{Z}^d$ are $\mathit adjacent$ if $|\lambda - \lambda^{\prime}| = 1$, where $|\cdot|$ is the $L^1$ metric. For each $T \subset \mathbb{Z}^d$, define $$\partial T = \{v \in \mathbb{Z}^d \setminus T \hspace{1mm}: \hspace{1mm} v \mbox{ is adjacent to some } \lambda \in T\}.$$
Let $\Lambda_n = \{\lambda \in \mathbb{Z}^d \hspace{1mm}: \hspace{1mm} ||\lambda|| < n\}$.
Claim. $|\partial \Lambda_n| = 2d(2n - 1)^{d-1}$
I could just justify that intuitively: the box $\Lambda_n$ has 2$d$ faces in $\mathbb{Z}^d$ and each of this faces have $(2n-1)^{d-1}$ adjacent vertices that are not in $\Lambda_n$.
How could I prove that more "mathematically"?
If $|\lambda-\lambda'|=1$ in $L^1$, that means that $\lambda$ and $\lambda'$ are equal in exactly $d-1$ components and differ in the remaining component by exactly $1$. That means for any $\lambda' \in \partial \Lambda_n$ that $\lVert\lambda'\rVert \le n$ holds. Hence,
$$\partial \Lambda_n \subseteq \{\lambda': \lVert\lambda'\rVert=n)$$
since all $\lambda'$ with smaller norm are in $\Lambda_n$.
A $\lambda' \in \partial \Lambda_n$ cannot have two or more components of absolute value equal to $n$, because any adjacent $\lambda$ would have at least one component of absolute value equal to $n$, contradicting the definition of $\partial \Lambda_n$. So we know that
$$\partial \Lambda_n \subseteq \{\lambda': \lVert\lambda'\rVert=n \;\land |\lambda'_i|=n \text{ for exactly one } i \in 1,2,\ldots,d \}$$
OTOH, each $\lambda' \in \{\lambda': \lVert\lambda'\rVert=n \;\land |\lambda'_i|=n \text{ for exactly one } i \in 1,2,\ldots,d \}$ has an adjacent $\lambda \in \Lambda_n$: Just change the one component $\lambda'_i$ with $|\lambda'_i|=n$ by $1$ in the direction of $0$ (this requires $n>0$), then for the resulting $\lambda$ we have $\lVert\lambda\rVert=n-1$.
This means that
$$\partial \Lambda_n = \{\lambda': \lVert\lambda'\rVert=n \;\land |\lambda'_i|=n \text{ for exactly one } i \in 1,2,\ldots,d \}.$$
Now the number of elements in the set on the right hand side is easy to count: Pick the component index $i$ with $|\lambda'_i|=n$ ($d$ choices), then choose the value for it (2 choices, $n$ or $-n$). Each other component (there are $d-1$ of them) can have independently any one of the values from $-(n-1)$ to $n-1$ ($2n-1$ choices).
The chosen index $i$ is uniquely determined, so any choices here lead to different elements of $\partial\Lambda_n$, which finally confirms your formula.