Yes, use the universal divisibility test: $ $ repeatedly replace leading digit chunks by their remainder mod the divisor as below, using least magnitude remainders $\, -8\le r \le 9\,$ to simplify arithmetic $\!\bmod 17\ $ (so negative digits occur, denoted $-d,c := 10(-d)+c\equiv 7d+c,\,$ by $10\equiv -7)$
$$\begin{align}
\!\bmod 17\!:\,\ 10\equiv -7\ \Rightarrow\quad\ \ &\ \color{#0A0}{1\ 2}\,\ 3\ 4\ 5\ 6\ \ \ {\rm by}\,\ \ \ \ \ \ \color{#0a0}{1\:\!2\equiv -5}\\[.1em]
\equiv\, &\ \color{#0A0}{{-5}},\color{#c00} 3\ 4\ 5\ 6\ \ \ {\rm by}\,\ \color{#0a0}{{-}5},\color{#c00}3\equiv\ (\,7\,)\,
\color{#0a0}{5}+\color{#c00}3\,\equiv\,\color{#0af} 4\\[.1em]
\equiv\, &\ \ \ \ \ \ \ \ \color{#0af}4\ 4\ 5\ 6\ \ \ {\rm by}\,\ \ \ \ \ \ \color{#0af}4\:\!4\equiv (-7)4+4\,\equiv\color{#f60}{-7}\\[.1em]
\equiv\, &\ \ \ \ \ \ \ \ \color{#f60}{{-}\!7}, 5\ 6\ \ \ {\rm by}\,\ \color{#f60}{{-}7},5\equiv\,(\,7\,)\, \color{#f60}{7}+5\,\equiv\, 3\\[.2em]
\equiv\, &\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ 6\ \ \ \:\!\:\!{\rm by}\ \ \ \ \ \ 3\:\! 6\equiv (-7)3+6\,\equiv\, 2\\
\equiv\, &\qquad\qquad\ 2;\ \ {\bf Quicker,}\,\ 2 \,\text{ digits at a time:}\\[.4em]
\!\bmod 17\!:\,\ 10^2\equiv -2\ \Rightarrow\quad\ \ &\ \color{#0A0}{12\ 3 4}\ 56\ \ \ {\rm by}\,\ \color{#0A0}{12\ 3 4}\equiv\, (-2)\color{#0a0}{\,12+34\equiv 10}\\[.1em]
\equiv\, &\ \ \ \ \ \ \color{#0A0}{{10}}\ \color{#c00}{56}\ \ \ {\rm by}\,\ \color{#0a0}{10}\ \color{#c00}{56}\equiv\ (-2)\,
\color{#0a0}{10}+\color{#c00}{56}\,\equiv\,\color{#0af}2\\[.2em]
\equiv\, &\qquad\quad\ \color{#0af}2
\end{align}\qquad\qquad$$
So $\rm\, 123456\equiv 2\pmod{\!17}.\,$ Indeed $\rm\, 123456 = 7262\cdot 17+2.\,$ Continuing this way we can do the entire number in a couple minutes of mental arithmetic. Unlike some other divisibility tests that compute only a binary truth value, this method has the advantage of computing the remainder. Further, it doesn't require remembering any special algorithm or parameters for each modulus.
Remark $ $ Lab & Steven's answers are a special case of above (but without mod arithmetic optimizations), i.e. they use chunk sizes of $\,2\,$ and $\,8,\,$ using $\bmod 17\!:\ 10^2\equiv -2,\ 10^8\equiv -1$.