What is the maximum value that can result from a 2D DCT?
As suggested in the comments by Marcus Müller, you have to start from the equation of the DCT:
$$
S_{uv} = \frac{1}{4} C_u C_v \sum_{y=0}^7 \sum_{x=0}^7 s_{xy} \cos\frac{(2x+1)u\pi}{16} \cos\frac{(2y+1)v\pi}{16}
$$
The argument of the summation is of the magnitude of $s_{xy}$ times two values which are at most 1 (absolute value). So still 8 bits. You are summing $8 \times 8 = 64$ of those values, so $2^6 \times 2^8 = 2^{14}$. But you also divide this by $\frac{1}{4}$ so the maximum value requires 12 bits. You also have to consider that this value is obtained only when u anv v are 0, and in that case $C_u=C_v=\frac{1}{\sqrt{2}}$, so $C_uC_v=\frac{1}{2}$, that is another bit is unneeded. Some other checks are required when only $u=0$ or $v=0$, but overall you will need 11 bits at most.
This is exactly seen in the JPEG category tables for encoding the DCT coefficients.
For the skeptics, as you can see in the table below, when the input is in [0,255] the maximum values are always below 2048, thus 11 bits are enough. If your input is in [-255,255] you will need 12 bits.





0
1
2
3
4
5
6
7




0
2040
924,25
942,3571
924,25
1020
924,25
942,3571
924,25


1
924,25
837,4883
853,8957
837,4883
924,25
837,4883
853,8957
837,4883


2
942,3571
853,8957
870,6245
853,8957
942,3571
853,8957
870,6245
853,8957


3
924,25
837,4883
853,8957
837,4883
924,25
837,4883
853,8957
837,4883


4
1020
924,25
942,3571
924,25
1020
924,25
942,3571
924,25


5
924,25
837,4883
853,8957
837,4883
924,25
837,4883
853,8957
837,4883


6
942,3571
853,8957
870,6245
853,8957
942,3571
853,8957
870,6245
853,8957


7
924,25
837,4883
853,8957
837,4883
924,25
837,4883
853,8957
837,4883