The answer is $\frac{4^k - 1}{3}$ if $3 \nmid k$, and $\frac{4^k + 2}{3}$ if $3 \mid k$.

Note that $n$ has exactly $k$ digits, possibly starting with $0$. We need to count the number $N$ of integers of the form $\overline{a_1a_2\cdots a_k}$ such that each $a_j$ is one of $0$, $1$, $2$, $7$ and $3 \mid a_1 + a_2 + \cdots + a_k$. Consider the generating function

$$f(x) = (1 + x + x^2 + x^7)^k = \sum_{\substack{a_1, \dots, a_k \\ a_j \in \{0, 1, 2, 7\}}} x^{a_1 + a_2 + \cdots + a_k}.$$

Let $c_j$ be the coefficient of $x^j$ for each $j$. Then $N = \sum_{3 \mid j} c_j$.

Let $\omega = \frac{-1}{2} + \frac{\sqrt{3}}{2}i$ be a cube root of unity. This means a number satisfying $\omega^3 = 1$ and $\omega \neq 1$. It has the property that $\omega^{2j} + \omega^j + 1 = 0$ if $3 \nmid j$, and $\omega^{2j} + \omega^j + 1 = 3$ if $3 \mid j$. It follows that

$$\begin{align*} 3N &= \sum_j c_j (1^j + \omega^j + \omega^{2j}) \\ &= f(1) + f(\omega) + f(\omega^2) \\ &= 4^k + (1 + \omega + \omega^2 + \omega^7)^k + (1 + \omega^2 + \omega^4 + \omega^{14})^k \\ &= 4^k + \omega^k + \omega^{2k}. \end{align*}$$

If $3 \nmid k$, we have $\omega^k + \omega^{2k} = -1$. This gives $N = \frac{4^k - 1}{3}$.

If $3 \mid k$, we have $\omega^k = \omega^{2k} = 1$. This gives $N = \frac{4^k + 2}{3}$.