You're nearly there.
A square matrix $B$ is non-invertible if and only if there exists a non-zero vector $v$ such that $Bv=0$. For example, necessity follows because $Bv=0$ implies that the function $f(v)=Bv$ is not injective (or "one-to-one") and hence not bijective (which is necessary for $f$ to have an inverse -- see the link). It is easy to see that $B$ is not injective since $B(2v)=2Bv=2\cdot0=0=Bv$, that is, $f$ maps both $v$ and $2v$ onto the same point $0$.
So, if $\lambda$ is an eigenvalue of $A$, and $x$ is its corresponding eigenvector,
$$Ax=\lambda x\Leftrightarrow Ax-\lambda x=0\Leftrightarrow (A-I\lambda)x=0.$$
Hence, $\lambda$ must be such that $B=A-I\lambda$ is non-invertible. Thus $\lambda$ is an eigenvalue of $A$ if and only if it satisfies the characteristic equation $\det(A-I\lambda)=0$.
Aside: If $\lambda$ is real, $x$ is simply a vector that function $f(v)=Av$ maps onto "itself" just stretches it and/or reflects it across the origin. For example, if $\lambda=2$, $f(x)$ simply "stretches" $x$ by two, and if $\lambda=-1$, $f(x)$ reflects $x$ across the origin (rotates it by $180$ degrees).
Edit: You might find these cam-casts of interest.
I'm going to actually give a couple of hints that should help you solve this.
First, you already have seen that 0 is an eigenvalue, probably by noticing that your columns are not linearly independent. Try to see if you can find the rank of the matrix by observation. That will let you find the multiplicity of 0 as an eigenvalue.
Your matrix has constant row sums, this row sum will be an eigenvalue (with the all-ones vector as an eigenvalue).
Next, recall that the trace of a matrix is the sum of the eigenvalues (including multiplicity). That should help you narrow things down as well.
Once you have the eigenvalues, you have that $A = P^{-1} D P$, where $D$ is a diagonal matrix consisting of the eigenvalues on the diagonal. So $A^{2017} = P^{-1} D^{2017} P$, you will only be able to do this in your head if $D$ is reaaaaly really nice looking (it will be in this case).
Best Answer
Take an eigen vector $v$ corresponding to an eigenvalue $\lambda$. Use this fact and cacluate $A^2v$ and $6Av$ independently, and equate them using the information $A^2=6A$; that will give you a condition on $\lambda$ enabling you to guess it.