Let $D \in \mathbb{R}^{n \times n}$ defined as
\begin{equation}
D := \begin{pmatrix}
1 & 0 & \cdots & \cdots & 0 \\
-1 & 1 & \ddots & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & 0 & \cdots & -1 & 1
\end{pmatrix}
\end{equation}
or in another form as
\begin{equation}
D_{i,j} := \begin{cases}
1 & \textrm{ if } i=j , \\
-1 & \textrm{ if } i=j+1 , \\
0 & \textrm{ otherwise }.
\end{cases}
\end{equation}
Then consider the matrix $L \in \mathbb{R}^{2n^{2} \times 2n^{2}}$ defined as
\begin{equation}
L := \left( d+1 \right) \begin{pmatrix}
\mathbb{I}_{n} \otimes D \\ D \otimes \mathbb{I}_{n}
\end{pmatrix} .
\end{equation}

> I would like to compute (or at least estimate) the smallest eigenvalue of $LL^{T}$.

The exact form of $LL^{T}$ can be computed. Indeed,
\begin{equation}
L^{T} := \left( d+1 \right) \begin{pmatrix}
\left( \mathbb{I}_{n} \otimes D \right) ^{T} , \left( D \otimes \mathbb{I}_{n} \right) ^{T}
\end{pmatrix} = \left( d+1 \right) \begin{pmatrix}
\mathbb{I}_{n} \otimes D^{T} , D^{T} \otimes \mathbb{I}_{n}
\end{pmatrix} .
\end{equation}
Thus
\begin{align}
LL^{T} & = \left( d+1 \right) ^{2} \begin{pmatrix}
\mathbb{I}_{n} \otimes D \\ D \otimes \mathbb{I}_{n}
\end{pmatrix} \begin{pmatrix}
\mathbb{I}_{n} \otimes D^{T} , D^{T} \otimes \mathbb{I}_{n}
\end{pmatrix} \\
& = \left( d+1 \right) ^{2} \begin{pmatrix}
\mathbb{I}_{n} \otimes DD^{T} & D^{T} \otimes D \\ D \otimes D^{T} & DD^{T} \otimes \mathbb{I}_{n}
\end{pmatrix} .
\end{align}
Probably we can use the Schur Complement to compute but it still too complicated. Since each of the above 4 blocks, the eigenvalues can be compute, I wonder is there any way to connect these eigenvalues with the one of $LL^{T}$.