$(1)$ Solution: From the given information, we can find $b_2= \frac {9}{2}$, $c_2= \frac {7}{2}$, $b_3= \frac {15}{4}$, $c_3= \frac {17}{4}$;
$(2)$ Solution: Since $a_{n+1}=a_n$ and $a_1=4$, we have $a_n=4$ $(n\in\mathbb{N}^*)$,
Therefore, $b_{n+1}= \frac {a_n+c_n}{2}= \frac {4+c_n}{2}= \frac {c_n}{2}+2$, $c_{n+1}= \frac {a_n+b_n}{2}= \frac {b_n}{2}+2$,
Then, $c_{n+1}-b_{n+1}= \frac {1}{2}(b_n-c_n)= -\frac {1}{2}(c_n-b_n)$,
Thus, the sequence $\{c_n-b_n\}$ is a geometric sequence with the first term $2$ and common ratio $- \frac {1}{2}$,
Therefore, $c_n-b_n=2\cdot (- \frac {1}{2})^{n-1}$;
$(3)$ Proof: From $(2)$, we know that $b_{n+1}+c_{n+1}= \frac {1}{2}(b_n+c_n)+4$,
Therefore, $b_{n+1}+c_{n+1}-8= \frac {b_n+c_n}{2}-4= \frac {1}{2}(b_n+c_n-8)$,
And since $b_1+c_1-8=0$,
Therefore, from the above recursive relationship, it can be concluded that when $n\in\mathbb{N}^*$, $b_n+c_n-8=0$ always holds,
That is, $b_n+c_n$ is a constant value of $\boxed{8}$.