Question: Does there exist a prime $p \geq 7$ such that the order of $4$ in the multiplicative group of units in $\mathbb{Z}/p^n$ is odd for every positive integer $n$? 

It would be nice if $7$ was already an example. I computed the order of $4$ modulo $7^n$ for $n=1,2,\ldots,12$ and it came out odd, but that's not particularly compelling evidence I guess.

Answer: 
Hint: $4$ is always a square modulo $p^n$. What can you say about the order of its square root (whatever that is) modulo $7^n$? What does this imply about the order of $4$?
Use Theorem 4.4 in Leveque, Fundamentals of Number Theory:

"Suppose $p$ is a prime and $p \nmid a$. Let $\mathrm{ord}\_p a = t$ and let $p^z$ be [the exact power of $p$ dividing $a^t-1$]. Then if $p>2$ or $z>1$,

$$ t\_n = \mathrm{ord}\_{p^n} a = t, \quad \text{for $n \leq z$}$$

and 

$$ t\_n = t p^{n-z}, \quad \text{for $n \geq z$}"$$

So if the order of $4$ modulo $p$ is odd, then the order of $4$ is odd modulo every power of $p$.