Let $V$ be the vector space of all sequences which are eventually zero. Let $L$ be the backwards shift-- this is obviously "locally nilpotent". Given $V$ the norm
$$ \| (x_n) \| = \sum_n a_n x_n, $$
where $(a_n)$ is some sequence of positive numbers. Let $e_n$ be the vector which is 1 in the $n$th place, and zero elsewhere. Then
$$ \exp(L)(e_n) = (\cdots,1/2,1,1,0,\cdots), $$
where the final 1 is in the $n$th place. So
$$ \|\exp(L)(e_n)\| / \|e_n\| \geq (a_{n-1}+a_n)/a_n = 1 + a_{n-1}/a_n. $$
Hence just choose $(a_n)$ so that $( a_{n-1}/a_n )$ is an unbounded sequence, and then $\exp(L)$ will be unbounded. E.g. $(a_n)=(1,2,1,3,1,4,1,\cdots)$ will work.

unbounded? – Yemon Choi Jul 21 '11 at 16:14