Defining the GHZ state
The (generalized) $n$-qubit GHZ state is the state
\begin{equation}
|\mathrm{GHZ}\rangle = \frac{1}{\sqrt{2}}\left(|0,0,\dots,0\rangle + |1,1,\dots,1\rangle\right).
\end{equation}
This state is a stabilizer state with generators
\begin{align}
X_{1}\dots X_{n} \\
Z_{1}Z_{2}\\
Z_{1}Z_{3}\\ \vdots \\
Z_{1}Z_{i+1} \\ \vdots \\
Z_{1}Z_{n}
\end{align}
for a total of $n$ generators $^{1}$.
However, the GHZ state is not a graph state, because it doesn't permit a set of 'canonical graph state generators' in the form of $X_{i} Z_{N_{i}}$ for every node in the graph (here, $N_{i}$ is the neighbourhood of node $i$).
Local equivalence to the star graph
The GHZ state is not a graph state, but it is local Clifford equivalent to a graph state$^{2}$. If you apply a Hadamard operation to all qubits except the first, the state becomes:
\begin{equation}
|\mathrm{GHZ}\rangle = \frac{1}{\sqrt{2}}\left(|0,+\dots,+,\rangle + |1,-,\dots,-\rangle\right).
\end{equation}
With generators:
\begin{align}
X_{1}\dots Z_{n} \\
Z_{1}X_{2}\\
Z_{1}X_{3}\\ \vdots \\
Z_{1}X_{i+1} \\ \vdots \\
Z_{1}X_{n}
\end{align}
This is a graph state, with node $1$ connected to all other nodes, and all other nodes connected only to node $1$. This is exactly the star graph with the first node being the connecting node!
We thus see that, while the GHZ state is not technically a graph state, it is local Clifford equivalent to the star graph. Therefore, often the star graph is assumed 'synonymous' with the GHZ state.
The complete graph
The complete graph is local Clifford equivalent to the star graph. One can show this by showing that, starting from the star graph around node $1$, applying the operation $\sqrt{X_{1}} \sqrt{Z_{2,\dots,n}}$ (i.e. a $\sqrt{X}$ on the first qubit, and $\sqrt{Z}$ on all other qubits) results in the graph state of the complete graph. However, actually performing this calculation is quite tedious.
Our life is made easier by the fact that the star and complete graph are related by a local complementation: a local complementation on the middle node in the star graph, results in the complete graph. Following the same paper I already linked, a local complementation on a graph is associated with a certain local Clifford operation on the graph state. Thus, since the star graph and the complete graph are related by a local complementation $\tau_{1}$, their associated graph states are related by a local Clifford operation (namely, the one I mentioned before).
Conclusion
So, in conclusion: The GHZ state, while not technically a graph state, is local equivalent to certain graph states: the star graph, and the complete graph. Note that the local complementation is invertible. Thus, if I locally complement on any other node (starting from the complete graph), I get a star graph around another middle node. All these graph states are locally equivalent to the GHZ state.
In my experience people associate both the star and the complete graph equally often with the GHZ state. Moreover, people might say the GHZ state 'is' the star graph or 'is' the complete graph, but this is technically not complete correct. However, in the study of quantum networks, these local operations don't really matter that much, and its easier to just work with graph states.
Studying the 'amount' of entanglement
You say that the GHZ state is supposed to be 'maximally entangled'; depending on your definition of what is maximally entangled this is indeed the case. But the local complementation operation shows a very important fact: just the number of edges in a graph state does not determine or even indicate the amount of entanglement! Local operations can easily change the number of edges, so they don't reflect the 'amount' of entanglement.
Bonus: action of the Hadamard on a graph state
The Hadamard operation is a local Clifford operation. However, consider applying a Hadamard operation to e.g. the first node of the star graph. The generators become:
\begin{align}
Z_{1}\dots Z_{n} \\
X_{1}X_{2}\\
X_{1}X_{3}\\ \vdots \\
X_{1}X_{i+1} \\ \vdots \\
X_{1}X_{n}
\end{align}
That is not a graph state, but 'just' a stabilizer state. In general, randomly chosen local Clifford operations applied to a graph state will result in a stabilizer state, and usually not a graph state.
However, remember: any stabilizer state is always local Clifford equivalent to a graph state! So again, in the study of entanglement in e.g. networks, we can forget about these local operations, and focus solely on graph states.
Notes
$^{1}$You might have seen generators of the form $Z_{i}Z_{i+1}$ instead of the first $Z$ always being on the first node. But they are equivalent by a simple change of generators, and this above form is easier to see the equivalence to the star graph.
$^{2}$In fact, every stabilizer state is local Clifford equivalent to a graph state, see e.g. section III of this paper. But this is currently besides the point.
