You can use the Cayley-Hamilton theorem, which says that the matrix $A$ is a root of the minimal polynomial, which divides the characteristic polynomial. In facts, the minimal polynomial has the same roots than the characteristic, except those roots may have lower multiplicity. In particular, if the eigenvalues of $A$ are distinct, then the two polynomials are equals.
You compute the $n=\dim$ powers: $I, A, A^2, A^3,...,A^n$, and then look for coefficients $c_i$ such that $A^n = c_{n-1}A^{n-1}+...+c_1A^1+c_0I$, then the minimal polynomial is $\det(\lambda I - A) = P(X) = \lambda^n-c_{n-1}\lambda^{n-1}-...-c_1\lambda-c_0$.
For example, if $A=\begin{pmatrix} 0&1&0\\1&0&0\\0&0&2\end{pmatrix}$, then $A^2=\begin{pmatrix} 1&0&1\\0&1&0\\0&0&4\end{pmatrix}$, and $A^3=\begin{pmatrix} 0&1&0\\1&0&0\\0&0&8\end{pmatrix}$,
so that $A^2-I=\begin{pmatrix} 0&0&0\\0&0&0\\0&0&3\end{pmatrix}$ and $A^3-A=\begin{pmatrix} 0&0&0\\0&0&0\\0&0&6\end{pmatrix}$, thus $A^3-A=2(A^2-I)$, and
the minimal polynomial is $\lambda^3-\lambda-2(\lambda^2-1)=(\lambda^2-1)(\lambda-2)=(\lambda+1)(\lambda-1)(\lambda-2)$. All the roots being simple, the minimal polynomial is also the characteristic polynomial, and $\chi_A(\lambda)=\det(\lambda I-A)=\lambda^3-2\lambda^2-\lambda+2$.
The search of the coefficients $c_i$ may be done in a systematic way, similar to the Gauss method. However the system is overdetermined, with $n \times n$ equations, and is usually quicker to solve than you think, especially for matrices found in homework.
The eigenvalues will almost always be distinct (so the characteristic and minimal polynomials will be the same), except maybe for matrices found in homework. For these, is is better to look to the power matrix $A^k, k \le n$ as you compute them, and look if they are linear combinations of the previous ones.
For example if $A=\begin{pmatrix} 0&1&1\\1&0&1\\1&1&0\end{pmatrix}$, then $A^2=\begin{pmatrix} 2&1&1\\1&2&1\\1&1&2\end{pmatrix}$, and $A^2=A+2I$. So the minimal polynomial is $\lambda^2-\lambda-2 = (\lambda+1)(\lambda-2)$. The characteristic polynomial being a polynomial of degree 3 with the same roots, it can either be $(\lambda+1)^2(\lambda-2)$ or $(\lambda+1)(\lambda-2)^2$. The multiplicity $\nu_i$ of $(x-\lambda_i)$ in $\chi_A(x) = \prod (x-\lambda_i)^{\nu_i}$, is the dimension of the associated eigenspace $E_{\lambda_i}=\ker (A-\lambda_i I)=\{x\mid Ax=\lambda_i x\}$. In our case, $E_{-1} = \ker(A+I) = \ker \begin{pmatrix} 1&1&1\\1&1&1\\1&1&1\end{pmatrix}=2$, so $\chi_A(\lambda)=(\lambda+1)^2(\lambda-2)$.
I don't know if the method is less tedious, but I find it less boring.