The answer is
$$f(n) = \begin{cases} 6 & \text{if } n = 1; \\ 4n + 1 & \text{if } n \ge 2. \end{cases}$$
We will use connected as a synonym for directly communicating, call a set of $k$ terminals for which each of the $\binom{k}{2}$ pairs of terminals is connected complete and call a set of $2k$ terminals forming $k$ disjoint connected pairs a k-matching.

We first show that $f(n) = 4n + 1$ for $n > 1$. The $4n$-terminal network consisting of two disconnected complete sets of $2n$ terminals clearly does not contain an $n$-bladed windmill (henceforth called an $n$-mill), since such a windmill requires a set of $2n+1$ connected terminals. So we need only demonstrate that $f(n) = 4n + 1$ is sufficient.

Note that we can inductively create a $k$-matching in any subnetwork of $2k + 1$ elements, as there is a connected pair in any set of three or more terminals. Also, the set of terminals that are not connected to a given terminal $x$ must be complete, as otherwise there would be a set of three mutually disconnected terminals. We now proceed by contradiction and assume that there is a $(4n + 1)$-terminal network without an $n$-mill. Any terminal $x$ must then be connected to at least $2n$ terminals, for otherwise there would be a complete set of size at least $2n + 1$, which includes an $n$-mill. In addition, $x$ cannot be directly connected to more than $2n$ terminals, for otherwise we could construct an $n$-matching among these, and therefore an $n$-mill. Therefore every terminal is connected to precisely $2n$ others.

If we take two terminals $u$ and $v$ that are not connected we can then note that at least one must be connected to the $4n - 1$ remaining terminals, and therefore there must be exactly one, $w$, to which both are connected. The rest of the network now consists of two complete sets of terminals $A$ and $B$ of size $2n - 1$, where every terminal in $A$ is connected to $u$ and not connected to $v$, and every terminal in $B$ is not connected to $u$ and connected to $v$. If $w$ were connected to any terminal in $A$ or $B$, it would form a blade with this element and hub $u$ or $v$ respectively, and we could fill out the rest of an $n$-mill with terminals in $A$ or $B$ respectively. Hence $w$ is only connected to two terminals, and therefore $n = 1$.

Examining the preceding proof, we can find the only 5-terminal network with no 1-mill: With terminals labeled A, B, C, D, and E, the connected pairs are (A, B), (B, C), (C, D), (D, E), and (E, A). (As indicated in the figure above, a pair of terminals are connected if and only if the edge connecting them are darkened.) To show that any 6-terminal network has a 1-mill, we note that any complete set of three terminals is a 1-mill. We again work by contradiction. Any terminal a would have to be connected to at least three others, b, c, and d, or the terminals not connected to a would form a 1-mill. But then one of the pairs (b, c), (c, d), and (b, d) must be connected, and this creates a 1-mill with that pair and a.