This is not a full answer, but let me point out that there is an explicitlish formula for the ternary expansion of such numbers, for any given value of $3$.

For integers $a$ and $m>0$, let $a\bmod m$ denote $a-m\lfloor a/m\rfloor$ (the unique residue in $\{0,\dots,m-1\}$ congruent to $a$ modulo $m$). Inside the left argument of this operator, $a^{-1}$ denotes multiplicative inverse of $a$ modulo $m$.

**Proposition:** Let $b,m,n\in\mathbb N$, where $m>1$ and $\gcd(b,m)=1$. Then
$$\def\FL{\genfrac\lfloor\rfloor{}{}}\FL{b^n}m=\sum_{i<n}b^i\bigl((-m^{-1}(b^{n-i}\bmod m))\bmod b\bigr).$$

**Proof:**
Using a telescoping sum,
$$\begin{align*}
\FL{b^n}m
&=\sum_{i<n}b^i\left(\FL{b^{n-i}}m-b\FL{b^{n-i-1}}m\right)\\
&=\sum_{i<n}b^i\left(\frac{b^{n-i}-(b^{n-i}\bmod m)}m-\frac{b^{n-i}-b(b^{n-i-1}\bmod m)}m\right)\\
&=\sum_{i<n}b^i\left(\frac{b(b^{n-i-1}\bmod m)-(b^{n-i}\bmod m)}m\right).
\end{align*}$$
Now, if we put $u=(b^{n-i-1}\bmod m)$ and $v=(b^{n-i}\bmod m)$, then $v=(bu\bmod m)$; that is, $bu-v=mr$, where $r=\lfloor bu/m\rfloor$ is the unique element of $\{0,\dots,b-1\}$ such that $b\mid v+mr$. In other words, $r$ is $(-m^{-1}v)\bmod b$. QED

**Corollary:** If $p\equiv\pm1\pmod 3$ is prime, then
$$\frac{3^{p-1}-1}p=\sum_{i<p-1}3^i\bigl((\mp(3^{p-1-i}\bmod p))\bmod 3\bigr).$$

This allows to answer some questions about the distribution of ternary digits in this number, for example:

If $p\equiv2\pmod3$, then digit $1$ appears at position $0$.

If $3$ is a primitive root modulo $p$, then $\{3^{p-1-i}\bmod p:i<p-1\}=\{1,\dots,p-1\}$, thus digit $1$ appears $(p\mp1)/3$ many times.

It’s tempting to think that “roughly $1/3$” of the digits should be $1$ in any case, but in the extreme example $p=3^k-1$ (which can’t be prime, though), digit $1$ has frequency $1/k$ and digit $2$ does not appear at all.