Find the maximum flow
Charcoal, 37 36 34 bytes
Ｉ⌊ＥＥＸ²⁻Ｌθ²↨⁺Ｘ²⊖Ｌθ⊗ι²ΣＥθ∧§ιμΣΦλ¬§ιξ

Try it online! Link is to verbose version of code. Explanation:
   Ｅ…Ｘ²⁻Ｌθ²Ｘ²⊖Ｌθ                    Loop over all cuts...
  Ｅ             ↨⊗ι²                ... converted to base 2
                    ΣＥθ∧§ιμ         Sum rows in source cut
                           ΣΦλ¬§ιξ  Sum columns in sink cut
Ｉ⌊                                  Print the minimum

The range is constructed so that the base 2 values are 100...000 to 111...110 whereby the 1 bits refer to the source cut (thus the left bit, which is actually element 0, is the source itself) and the 0 bits refer to the sink cut.