Assuming that the discussion is restricted to single-qubit states then the answer to your question is no. The unambiguous discrimination of non-orthogonal states relies on using POVM elements that are proportional to elements of the orthogonal complements of the states in question:
$$
E_1 \propto |\phi_1^{\perp}\rangle\langle\phi_1^{\perp}|, \qquad
\langle\phi_1^{\perp}|\phi_1\rangle = 0,
$$
and so on. This choice of POVM element ensures that the input state $|\phi_1\rangle$ will never yield the outcome '1', and observing this measurement outcome therefore reveals the information that the input state wasn't $|\phi_1\rangle$. If you know that the input state belongs to the set $\{|\phi_1\rangle,|\phi_2\rangle\}$ this in turn implies that the state was $|\phi_2\rangle$.
Extending this to more than two non-orthogonal single-qubit states would require a POVM element that is proportional to an element of the union of the orthogonal complements of two different states. In other words:
$$
E \propto |\phi^{\perp}\rangle\langle\phi^{\perp}|,
\qquad
\langle\phi^{\perp}|\phi_1\rangle = 0,
\quad \langle\phi^{\perp}|\phi_2\rangle = 0,
$$
but for qubits this implies that $|\phi_1\rangle = |\phi_2\rangle$ since for every single-qubit state there is exactly one orthogonal single-qubit state.
In general, when considering states of more qubits, a necessary and sufficient condition for being able to discriminate $n$ non-orthogonal states is that they are linearly independent. See arXiv:quant-ph/9807022 for a reference.