Generalization of Schmidt decomposition
The Schmidt decomposition $|\psi_{AB}\rangle = \sum_i\lambda_i|i_A\rangle|i_B\rangle$, with $\lambda_i$ positive real numbers and $|i_A\rangle$ and $|i_B\rangle$ (possibly incomplete) orthonormal bases, of a bipartite state $|\psi_{AB}\rangle$ can be thought of as a type of singular-value decomposition. Specifically, if the amplitudes $\psi_{ij} = \langle i|\langle j|\psi_{AB}\rangle$ are arranged into a $2\times 2$ matrix rather than the usual $4$-vector, then $\lambda_i$ are precisely the non-zero singular values of the matrix $\psi_{ij}$. This explains why $\lambda_i$ are positive real numbers. See the proof of theorem 2.7 on page 109 in Nielsen & Chuang for more details.
The decomposition can be generalized to $n$-partite state as
$$
|\psi_{A_1A_2 \dots A_n}\rangle = \sum_i \lambda_i |i_{A_1}\rangle |i_{A_2}\rangle\dots |i_{A_n}\rangle\tag1
$$
where as before $\lambda_i$ are positive real numbers and $|i_{A_k}\rangle$ are (possibly incomplete) orthonormal bases. If $n=2$, then decomposition $(1)$ exists for every pure state $|\psi_{A_1A_2}\rangle$. If $n>2$, then there are pure states without decomposition $(1)$.
Physical characterization
Physically, the $n$-partite states that admit decomposition $(1)$ can be characterized as the states that are free of $m$-partite entanglement for every $1<m<n$. More precisely, $|\psi_{A_1A_2 \dots A_n}\rangle$ admits decomposition $(1)$ if and only if for every $k = 1,\dots n-1$ and subsystems $A_{i_1}\dots A_{i_k}$, the state
$$
\rho_{A_{i_1}\dots A_{i_k}} = \mathrm{tr}_{A_{i_1}\dots A_{i_k}}\left(|\psi_{A_1A_2 \dots A_n}\rangle\langle\psi_{A_1A_2 \dots A_n}|\right)
$$
is separable. One way of thinking about these states is that all (if any) entanglement that they have lives between all $n$ subsystems. Conversely, states that do not admit $(1)$ necessarily contain some entanglement between fewer than $n$ subsystems.
This characterization is not very useful to computationally check whether a given $n$-partite state admits decomposition $(1)$. Nevertheless, it helps to intuitively understand the class of Schmidt-decomposable states. In particular, it explains why a generic $n$-partite state for $n>2$ does not admit $(1)$. It also explains why all bipartite states admit $(1)$ $-$ it is impossible to entangle fewer than $2$ subsystems.
Necessary and sufficient conditions in the tripartite case
The necessary and sufficient conditions for a state to admit decomposition $(1)$ are given in the paper cited in a comment above by @Rammus. We reproduce the results of the paper in the case when the Schmidt decomposition
$$
|\psi_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle\tag2
$$
of a tripartite state $|\psi_{ABC}\rangle$ associated with the partitioning of $ABC$ into $A$ and $BC$ has distinct coefficients, i.e. $\lambda_i \ne \lambda_j$ for $i \ne j$. Define the matrices $\Omega_i$ as
$$
\Omega_{i,jk} = \langle j_B|\langle k_C|i_{BC}\rangle.
$$
In other words, $\Omega_i$ is the matrix of amplitudes of the state $|i_{BC}\rangle$. The necessary conditions for $|\psi_{ABC}\rangle$ to admit decomposition $(1)$ is for all $\Omega_i$ to be rank one and $\Omega_i^\dagger\Omega_{i'} = 0$ and $\Omega_i\Omega_{i'}^\dagger = 0$ for $i\ne i'$.
If $\lambda_i$ are not all distinct, then the necessary and sufficient conditions become more complicated, because $(2)$ and thus $\Omega_i$ are no longer unique. See the paper for more details.
Example: $W$ state
The $W$ state is
$$
|W\rangle = \frac{|001\rangle + |010\rangle + |100\rangle}{\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}}|0\rangle\otimes\frac{|01\rangle + |10\rangle}{\sqrt{2}} + \frac{1}{\sqrt{3}}|1\rangle\otimes |00\rangle.
$$
Thus, $\lambda_0=\frac{\sqrt{2}}{\sqrt{3}}$, $\lambda_1 = \frac{1}{\sqrt{3}}$, $|0_{BC}\rangle = (|01\rangle+|10\rangle)/\sqrt{2}$ and $|1_{BC}\rangle = |00\rangle$ so
$$
\Omega_0 = \frac{1}{\sqrt{2}}\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} \quad \Omega_1 = \begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}
$$
and we see that $\Omega_1$ is rank one, but $\Omega_0$ is not. Therefore, $W$ does not admit decomposition $(1)$.
