Yes, Alice and Bob can do better than option 1. However, Alice needs to send 2 classical bits to Bob. This is more than in option 2, but is still independent of $n$.
The idea is that instead of performing a measurement Alice applies a unitary to rotate the measured subspace onto an auxiliary qubit and then teleports the auxiliary qubit to Bob who performs the measurement.
Note that option 2 only succeeds in transmitting measurement outcome and not the post-measurement state, so we assume that this is the goal of the protocol.
Alice's unitary
Suppose Alice has an $n$-qubit state $|\psi_R\rangle$ and let $P_0$ and $P_1$ be the measurement projectors. By definition, the projectors satisfy the completeness and orthogonality relations
$$
P_0 + P_1 = I\\
P_0P_1 = 0.\tag1
$$
Define the $(n+1)$-qubit operator $U$ acting on the input register $R$ and an auxiliary qubit $M$ as
$$
U |\psi_R\rangle|0_M\rangle = P_0|\psi_R\rangle|0_M\rangle + P_1|\psi_R\rangle|1_M\rangle \\
U |\psi_R\rangle|1_M\rangle = P_0|\psi_R\rangle|1_M\rangle + P_1|\psi_R\rangle|0_M\rangle
$$
or in block matrix form
$$
U = \begin{pmatrix}
P_0 & P_1\\
P_1 & P_0
\end{pmatrix}.
$$
Using $(1)$ it is easily checked that $U$ is unitary.
Protocol
Alice prepares an auxiliary qubit in the $|0_M\rangle$ state and applies the $U$ to $|\psi_R\rangle|0_M\rangle$. Next, she teleports the state of the auxiliary qubit $M$ to Bob at the cost of transmitting two classical bits and consuming a Bell pair $(|0_A0_B\rangle + |1_A1_B\rangle)/\sqrt{2}$ where $A$ is the Alice's half of the pair and $B$ is the Bob's half. Note that teleportation preserves entanglement, so after teleportation the joint state of $R$ and $B$ is
$$
|\gamma_{RB}\rangle = P_0|\psi_R\rangle|0_B\rangle + P_1|\psi_R\rangle|1_B\rangle.
$$
Finally, Bob measures $B$ in computational basis.
Correctness
We need to prove that measuring $P_0$ and $P_1$ on the input state $|\psi_R\rangle$ has the same output statistics as measuring the post-teleportation state of $B$ in computational basis. The probability of the $0$ outcome in the former case is
$$
p(0|\psi_R) = \langle\psi_R|P_0|\psi_R\rangle
$$
and in the latter case it is
$$
p(0|\rho_B) = \langle 0|\rho_B|0\rangle = \langle\psi_R|P_0|\psi_R\rangle
$$
where
$$
\rho_B = \mathrm{tr}_R(|\gamma_{RB}\rangle\langle\gamma_{RB}|) = \langle\psi_R|P_0|\psi_R\rangle|0_B\rangle\langle 0_B| + \langle\psi_R|P_1|\psi_R\rangle|1_B\rangle\langle 1_B|
$$
is the post-teleportation state of $B$.
Teleportation preserves entanglement
Teleportation is often described as a protocol on three qubits $B$, $C$ and $D$ with $B$ in an arbitrary single-qubit state $|\psi_B\rangle = \alpha|0_B\rangle + \beta|1_B\rangle$ and $CD$ in the Bell state $(|0_C0_D\rangle + |1_C1_D\rangle)/\sqrt{2}$ at the end of which the qubit $D$ is in the state $|\psi_D\rangle = \alpha|0_D\rangle + \beta|1_D\rangle$.
Suppose that the input qubit $B$ is entangled with a qubit $A$ in a joint state
$$
|\phi_{AB}\rangle = \alpha|0_A0_B\rangle + \beta|0_A1_B\rangle + \gamma|1_A0_B\rangle + \delta|1_A1_B\rangle.
$$
We will show that at the end of the standard teleportation protocol the qubits $AD$ are in the state
$$
|\phi_{AD}\rangle = \alpha|0_A0_D\rangle + \beta|0_A1_D\rangle + \gamma|1_A0_D\rangle + \delta|1_A1_D\rangle
$$
thus proving that any entanglement the input qubit $B$ has with another system $A$ is preserved by the teleportation channel.
The initial state of the composite system $ABCD$ is
$$
\begin{align}
\frac{1}{\sqrt{2}} (&\alpha|00\rangle(|00\rangle +|11\rangle)\, + \\
& \beta|01\rangle(|00\rangle +|11\rangle)\, +\\
& \gamma|10\rangle(|00\rangle +|11\rangle)\, +\\
& \delta|11\rangle(|00\rangle +|11\rangle))
\end{align}
$$
where we have elided subsystem labels for legibility. After the CNOT gate with $B$ as the control and $C$ as the target and the Hadamard gate on $B$, the state of $ABCD$ becomes
$$
\begin{align}
& \frac{1}{2}(\alpha|0000\rangle + \beta|0001\rangle + \gamma|1000\rangle + \delta|1001\rangle)\, + \\
& \frac{1}{2}(\alpha|0011\rangle + \beta|0010\rangle + \gamma|1011\rangle + \delta|1010\rangle)\, + \\
& \frac{1}{2}(\alpha|0100\rangle - \beta|0101\rangle + \gamma|1100\rangle - \delta|1101\rangle)\, + \\
& \frac{1}{2}(\alpha|0111\rangle - \beta|0110\rangle + \gamma|1111\rangle - \delta|1110\rangle)
\end{align}
$$
where we have grouped the terms according to the state of the $BC$ subsystem. It is immediate that if measurement of $BC$ yields $|00\rangle$ then $AD$ collapses to $|\phi_{AD}\rangle$ as promised. Similarly, we see that the usual $X$ and $Z$ corrections on $D$ conditioned on the measurement results transform each of the other three groups into $|\phi_{AD}\rangle$ as well. $\square$
