Computing 'Robustness of Magic' of $n$-bit W states
Here is a proof that $R(W_n) \in \Omega(\sqrt[4]{n})$.

Given a $W_n$ state, it is possible to create $\lg n$ T states with Clifford operations and a single additional T gate. Basically:


Xor together all the qubit's whose binary index has an odd number of 1s, apply a T gate to that xor'd value, then unxor.
Apply an S gate to every qubit whose binary index has a number of 1s that's 2 or 3 more than a multiple of 4.
Apply a Z gate to every qubit whose binary index has a number of 1s that's 4, 5, 6, or 7 more than a multiple of 8.
Compress the unary $W_n$ state down into a binary register.


Here's an example circuit turning $W_8$ into 3 $|T\rangle$ states:



I also tested turning $W_{16}$ into 4 $|T\rangle$ states.

We can produce $\lg n$ T states from a $W_n$ state and a single $T$ state. It must always be the case that an output's robustness of magic is less than or equal than the input's. Therefore:

$$R(W_n) \cdot R(T) \geq R(W_n \otimes T) \geq R(T^{\otimes \lg n})$$

In the paper "Application of a resource theory for magic states to fault-tolerant quantum computing" it is proven (see page 3) that:

$$R(T^{\otimes k}) \in \Omega(1.2^k)$$

Meaning $R(W_n) \cdot R(T)$ is greater than or equal to something asymptotically lowerbounded by $1.2^{\lg n}$. Since $R(T)$ is a constant, we can ignore it and deduce:

$$\begin{align}
R(W_n) &amp;\in Ω(1.2^{\lg n})
       \\&amp;= Ω(n^{\lg 1.2})
       \\&amp;\subset Ω(n^{0.263})
       \\&amp;\subset Ω(\sqrt[4]{n})
\end{align}$$

Therefore $R(W_n)$ is asymptotically somewhere between $\sqrt[4]{n}$ (by synthesizing into Ts) and $n$ (via the decomposition I showed in the question).