TL;DR: It depends on how you choose to measure entanglement on a pair of qubits. If you trace out the extra qubits, then "No". If you measure the qubits (with the freedom to choose the optimal measurement basis), then "Yes".
Let $|\Psi\rangle$ be a pure quantum state of 3 qubits, labelled A, B and C. We will say that A and B are entangled if $\rho_{AB}=\text{Tr}_C(|\Psi\rangle\langle\Psi|)$ is not positive under the action of the partial transpose map. This is a necessary and sufficient condition for detecting entanglement in a two-qubit system. The partial trace formalism is equivalent to measuring qubit C in an arbitrary basis and discarding the result.
There's a class of counter-examples that show that entanglement is not transitive, of the form
$$
|\Psi\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|1\phi\phi\rangle),
$$
provided $|\phi\rangle\neq |0\rangle,|1\rangle$. If you trace out qubit $B$ or qubit $C$, you'll get the same density matrix both times:
$$
\rho_{AC}=\rho_{AB}=\frac12\left(|00\rangle\langle 00|+|1\phi\rangle\langle 1\phi|+|00\rangle\langle 1\phi|\langle\phi|0\rangle+|1\phi\rangle\langle 00|\langle0|\phi\rangle\right)
$$
You can take the partial transpose of this (taking it on the first system is the cleanest):
$$
\rho^{PT}=\frac12\left(|00\rangle\langle 00|+|1\phi\rangle\langle 1\phi|+|10\rangle\langle 0\phi|\langle\phi|0\rangle+|0\phi\rangle\langle 10|\langle0|\phi\rangle\right)
$$
Now take the determinant (which is equal to the product of the eigenvalues). You get
$$
\text{det}(\rho^{PT})=-\frac{1}{16}|\langle 0|\phi\rangle|^2(1-|\langle 0|\phi\rangle|^2)^2,
$$
which is negative, so there must be a negative eigenvalue. Thus, $(AB)$ and $(AC)$ are entangled pairs. Meanwhile
$$
\rho_{BC}=\frac12(|00\rangle\langle 00|+|\phi\phi\rangle\langle\phi\phi |).
$$
Since this is a valid density matrix, it is non-negative. However, the partial transpose is just equal to itself. So, there are no negative eigenvalues and $(BC)$ is not entangled.
Localizable Entanglement
One might, instead, talk about the localizable entanglement. Before further clarification, this is what I thought the OP was referring to. In this case, instead of tracing out a qubit, one can measure it in a basis of your choice, and calculate the results separately for each measurement outcome. (There is later some averaging process, but that will be irrelevant to us here.) In this case, my response is specifically about pure states, not mixed states.
The key here is that there are different classes of entangled state. For 3 qubits, there are 6 different types of pure state:
- a fully separable state
- 3 types where there is an entangled state between two parties, and a separable state on the third
- a W-state
- a GHZ state
Any type of quantum state can be converted into one of the standard representatives of each class just by local measurements and classical communication between the parties. Note that the conditions of $(q_1,q_2)$ and $(q_2,q_3)$ being entangled remove the first 4 cases, so we only have to consider the last 2 cases, W-state and GHZ-state. Both representatives are symmetric under exchange of the particles:
$$
|W\rangle=\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle)\qquad |GHZ\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)
$$
(i.e. if I swap qubits A and B, I still have the same state).
So, these representatives must have the required transitivity properties: If A and B are entangled, then B and C are entangled, as are A and C. In particular, Both of these representatives can be measured in the X basis in order to localize the entanglement. Thus, any pure state that you're given must be such that you can include the measurement to convert it into the standard representative into the measurement for localizing the entanglement, and you're done!
