First, an important distinction must be done. A Tucker decomposition of a third-order tensor $\mathcal{T}$ is any decomposition of the form
$$
\mathcal{T} = \mathcal{G} \times_1 \mathbf{A} \times_2 \mathbf{B} \times_3 \mathbf{C},
$$
where $\mathcal{G}$ is also a third-order tensor called core and $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are the matrix factors of the decomposition. In scalar form,
$$
t_{i_1,i_2,i_3} = \sum_{r_1=1}^{I_1} \sum_{r_2=1}^{I_2} \sum_{r_3=1}^{I_3} g_{r_1,r_2,r_3} a_{i_1,r_1} b_{i_2,r_2} c_{i_3,r_3}.
$$
where $I_1, I_2, I_3$ are the dimensions of $\mathcal{T}$. The flat (horizontal) matrix unfoldings of $\mathcal{T}$ are then written as
$$
\mathbf{T}_1 = \mathbf{A} \mathbf{G}_1 (\mathbf{C} \otimes \mathbf{B})^T \\
\mathbf{T}_2 = \mathbf{B} \mathbf{G}_2 (\mathbf{A} \otimes \mathbf{C})^T \\
\mathbf{T}_3 = \mathbf{C} \mathbf{G}_3 (\mathbf{B} \otimes \mathbf{A})^T,
$$
where $\otimes$ denotes the Kronecker product.
When the factor matrices are the matrices of left singular vectors of the singular value decompositions of each matrix unfolding, then we have a particular Tucker decomposition which is usually called the higher-order singular value decomposition (HOSVD) (see the seminal paper by De Lathauwer et al., ftp://ftp.esat.kuleuven.be/pub/SISTA/delathauwer/reports/ldl-94-31.pdf). Mathematically, expressing the SVDs of $\mathbf{T}_1$, $\mathbf{T}_2$ and $\mathbf{T}_3$ as
$$
\mathbf{T}_n = \mathbf{U}^{(n)} \mathbf{\Sigma}^{(n)} {\mathbf{V}^{(n)}}^H, \quad n = 1,2,3,
$$
the HOSVD of $\mathcal{T}$ is given by
$$
\mathcal{T} = \mathcal{S} \times_1 \mathbf{U}^{(1)} \times_2 \mathbf{U}^{(2)} \times_3 \mathbf{U}^{(3)},
$$
where the core satisfies
$$
\mathcal{S} = \mathcal{T} \times_1 {\mathbf{U}^{(1)}}^H \times_2 {\mathbf{U}^{(2)} }^H \times_3 {\mathbf{U}^{(3)}}^H.
$$
It is easy to verify the above formula for the core, as
$$
\mathbf{S}_1 = {\mathbf{U}^{(1)}}^H \mathbf{T}_1 ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T \\
= {\mathbf{U}^{(1)}}^H \mathbf{U}^{(1)} \mathbf{S}_1 ({\mathbf{U}^{(3)}} \otimes {\mathbf{U}^{(2)}})^T ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T = \mathbf{S}_1
$$
because each $\mathbf{U}^{(n)}$ is unitary by definition.
Now, the answer to your question can be immediately taken from the above formulas, since substituting the SVD of $\mathbf{T}_1$ in the expression for $\mathbf{S}_1$ we have
$$
\mathbf{S}_1 = {\mathbf{U}^{(1)}}^H \mathbf{T}_1 ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T
= {\mathbf{U}^{(1)}}^H \mathbf{U}^{(1)} \mathbf{\Sigma}^{(1)} {\mathbf{V}^{(1)}}^H ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T
= \mathbf{\Sigma}^{(1)} {\mathbf{W}^{(1)}}^H
$$
where we have defined
$$
{\mathbf{W}^{(1)}} = ({\mathbf{U}^{(3)}} \otimes {\mathbf{U}^{(2)}})^T {\mathbf{V}^{(1)}}.
$$
Note that
$$
{\mathbf{W}^{(1)}}^H {\mathbf{W}^{(1)}} = {\mathbf{V}^{(1)}}^H ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T ({\mathbf{U}^{(3)}} \otimes {\mathbf{U}^{(2)}})^T {\mathbf{V}^{(1)}} \\
= {\mathbf{V}^{(1)}}^H [({\mathbf{U}^{(3)}} \otimes {\mathbf{U}^{(2)}}) ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)]^T {\mathbf{V}^{(1)}} \\
= {\mathbf{V}^{(1)}}^H {\mathbf{V}^{(1)}} = \mathbf{I},
$$
and hence $\mathbf{W}^{(1)}$ is unitary. This tells us that the mode-1 unfolding of the core $\mathcal{S}$ is a product of the diagonal matrix of singular values of $\mathbf{T}_1$ by a unitary matrix. Therefore, the norm of the $i$th row of $\mathbf{S}_1$ is precisely the $i$th singular value $\sigma_i^{(1)}$ of $\mathbf{T}_1$. An analogous reasoning applies of course to the other modes.
Finally, observe also that
$$ \|\mathcal{S}\|_F^2 = \|\mathbf{S}_n\|_F^2 = \text{Trace}(\mathbf{S}_n\mathbf{S}_n^H) = \sum_i (\sigma_i^{(n)})^2 = \|\mathbf{T}_n\|_F^2 = \|\mathcal{T}\|_F^2,$$
which comes as no surprise, since $\mathcal{S}$ is given by a unitary multilinear transformation of $\mathcal{T}$.
