First, we will denote the $n^2\times n$ matrix of third-order derivatives as $\mathbf{K}$, which has the following structure:
$$
\mathbf{K}=\frac{\partial\operatorname{vec}(\mathbf{H})}{\partial\mathbf{x}^T}=\begin{bmatrix}
\frac{\partial f}{\partial \mathbf{x}_1\mathbf{x}_1\mathbf{x}_1} & \cdots & \frac{\partial f}{\partial \mathbf{x}_1\mathbf{x}_1\mathbf{x}_{n-1}} & \frac{\partial f}{\partial\mathbf{x}_1\mathbf{x}_1\mathbf{x}_n} \\
\frac{\partial f}{\partial \mathbf{x}_1\mathbf{x}_2\mathbf{x}_1} & \cdots & \frac{\partial f}{\partial \mathbf{x}_1\mathbf{x}_2\mathbf{x}_{n-1}} & \frac{\partial f}{\partial \mathbf{x}_1\mathbf{x}_2\mathbf{x}_n} \\
\vdots & \ddots & \vdots & \vdots \\
\frac{\partial f}{\partial \mathbf{x}_n\mathbf{x}_n\mathbf{x}_1} & \cdots & \frac{\partial f}{\partial\mathbf{x}_n\mathbf{x}_n\mathbf{x}_{n-1}} & \frac{\partial f}{\partial \mathbf{x}_n\mathbf{x}_n\mathbf{x}_n}
\end{bmatrix}
$$
and we note that the tri-linear operator is defined as follows:
$$
\nabla^3f(\mathbf{x})[\mathbf{a},\mathbf{b},\mathbf{c}]=\sum_{i,j,k}^n\frac{\partial f}{\partial\mathbf{x}_i\partial\mathbf{x}_j\partial\mathbf{x}_k}\mathbf{a}_{i}\mathbf{b}_{j}\mathbf{c}_{k}
$$

We will show the following relationship holds:
$$
\|\nabla^3f(\mathbf{x})\|\leq\|\mathbf{K}\|\leq \sqrt{n}\|\nabla^3f(\mathbf{x})\|
$$

**Proof**:

I claim that $\nabla^3f(\mathbf{x})[\mathbf{a},\mathbf{b},\mathbf{c}]=(\mathbf{a}\otimes \mathbf{b})^T\mathbf{K}\mathbf{c}$ -- that is to say $\mathbf{K}$ can perform the same action as the tri-linear operator. This can be seen by simply writing out the RHS and comparing it to the form of the tri-linear operator. Taking the absolute value and maximizing over all vectors with norm less than 1 gives us the following:
$$ \|\nabla^3f(\mathbf{x})\|=\max_{\|\mathbf{a},\mathbf{b},\mathbf{c}\|\leq 1}|\nabla^3f(\mathbf{x})[\mathbf{a},\mathbf{b},\mathbf{c}]|=\max_{\|\mathbf{a},\mathbf{b},\mathbf{c}\|\leq 1}|(\mathbf{a}\otimes \mathbf{b})^T\mathbf{K}\mathbf{c}|
$$
This immediately implies the LHS of the bound in question, as the RHS of the preceding equation is bounded above by $\|\mathbf{K}\|$. Now we can bound it below by choosing appropriate vectors. In particular, take $\mathbf{c}$ to be the left-most right singular vector of $\mathbf{K}$, and let $\mathbf{u}$ be the left-most left singular vector of $\mathbf{K}$. For now, we leave $\mathbf{a}$ and $\mathbf{b}$ unspecified. Computing the matrix-vector product $\mathbf{K}\mathbf{c}$ gives the following result:
$$
\|\mathbf{K}\|\cdot|(\mathbf{a}\otimes \mathbf{b})^T\mathbf{u}|=|(\mathbf{a}\otimes \mathbf{b})^T\sigma_{1}\mathbf{u}|\leq\|\nabla^3f(\mathbf{x})\|
$$
where $\sigma_1$ is the largest singular value of $\mathbf{K}$. Now, we will rewrite the LHS as $\|\mathbf{K}\|\cdot|\mathbf{a}^T\operatorname{mat}(\mathbf{u})\mathbf{b}|$. We know that $\|\operatorname{mat}(\mathbf{u})\|_F=1$ as the vector $\mathbf{u}$ has unit norm. Thus we can bound the largest singular vector of $\operatorname{mat}(\mathbf{u})$ below by $1/\sqrt{n}$.

Then let $\mathbf{b}$ be the left-most right singular vector of $\operatorname{mat}(\mathbf{u})$, and let $\mathbf{a}$ be the left-most left singular vector of the same matrix. We can now write the following:
$$
\|\mathbf{K}\|\cdot|\mathbf{a}^T\operatorname{mat}(\mathbf{u})\mathbf{b}|\geq\frac{1}{\sqrt{n}}\|\mathbf{K}\|
$$
Rearranging gives us the upper bound of the result in question.