Why does the NIST CTR DRBG apparently output the internal state when no additional input is given?
Still, the new internal state starts with 
  
  $E(K,V+1) || E(K,V+2) || ... $. 


There is where you are incorrect; the CTR_DRBG_GENERATE function updates $V$ while it runs, and hence the new internal state is actually>

$E(K,V+n+1) || E(K,V+n+2) || ... $

(where the $V$ value is initial value of $V$ when generate is called, not when the update is called).

Because these are encryptions of blocks that are distinct from the blocks used to generate the random output, there is minimal leakage of the new state (well, someone looking at the random output could deduce what the new state is not; however that's not of much help to an attacker)