I’m sure that you meant to say that the number was formed by $3^n$ digits equal to $1$, not $3^3$ digits.
Let $m_n$ be the number whose decimal representation is a string of $3^n$ $1$’s. To be explicit, $m_n=\frac19(10^{3^n}-1)$. Suppose that $3^n$ divides $m_n$ for some $n$, say $m_n=3^nk$. Then $\begin{align*} m_{n+1}&=10^{3^n}\left(10^{3^n}m_n+m_n\right)+m_n\tag{1}\\ &=\left(10^{2\cdot3^n}+10^{3^n}+1\right)m_n\\ &=\left(10^{2\cdot3^n}+10^{3^n}+1\right)3^nk\;, \end{align*}$
and we need only show that $3$ divides $10^{2\cdot3^n}+10^{3^n}+1$. But $10\equiv 1\pmod 3$, so every power of $10$ is congruent to $1$ mod $3$, and therefore $10^{2\cdot3^n}+10^{3^n}+1$ is indeed a multiple of $3$.
If the step $(1)$ is a little puzzling at first, look at an example:
$m_2=111111111=111000000+111000+111=1000(1000m_1+m_1)+m_1\;.$