Water held in a hexagonal rod scuplture
Python 2, 222 bytes

import sys
y=h=v=0;B={}
for l in sys.stdin:
 z=y;y+=2j
 while l:
    if"0"&lt;l:B[z]=int(l[:2])
    l=l[2:];z+=1
while B:C=B;B={b:B[b]for b in B if(h&lt;B[b])+sum(3&gt;abs(c-b)for c in B)/7};a=C==B;h+=a;v+=a*sum(h&gt;B[b]for b in B)
print v


Reads input through STDIN and writes the result to STDOUT.

Explanation

We start at zero and incrementally increase the water level as follows:
Suppose that the water level is h, and we want to add 1 centimeter of water.
We'll call hexagons of height h or less, ones that are about to go (or already are) underwater, "submerged."
The water will spill through any submerged hexagon that isn't surrounded by six neighbors.
We eliminate all such hexagons; of course, now some other submerged hexagons might have less than six neighbors, and they need to be eliminated as well.
We continue in this fashion until convergence, i.e., until all remaining submerged hexagons have exactly six neighbors.
At this point we add the number of submerged hexagons (the gained water volume) to the total count, and increment the water level.

Eventually, all the hexagons will have been eliminated and we halt.