Suppose that $H^0(L^{\otimes n})|_ Y$
is non-zero for all $Y$ and $n$ sufficiently big, and vanishes somewhere on $Y$. Then it follows that the base
set of $L$ is trivial: indeed, $L$ has a non-zero
section on any complex subvariety, which includes
the base set. This implies that the natural map
$P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$
is holomorphic, for $n$ sufficiently big.
Also from your assumption it follows that
$P_n$ does not map any irreducible, positive-dimensional
subvariety to a point (again, for $n$ sufficiently big).
This implies that $P_n$ is a finite, proper map
to a projective variety, hence $X$ is a
ramified covering of a projective variety.
A ramified covering of a projective variety
is projective, which can be seen from vanishing
of cohomology of powers of $L$ (a finite
map is acyclic on coherent sheaves, hence
the cohomology of $L^{\otimes n}$ on $X$ are the
same as cohomology of ${\cal O}(1)$
on its image).

However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$;
the implication $H^0(L^{\otimes n} |_Y)\neq 0$
$\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy.

If $Y$ does not lie in a zero divisor of $L^{\otimes n}$,
we are done,  otherwise we replace $Y$ with the zero
divisor of $L^{\otimes n}$. Using induction on dimension,
we may already assume that the restriction $L|_Y$ is ample.
Consider the exact sequence
$$
0\rightarrow H^0(L^{\otimes n+k})
\rightarrow H^0(L^{\otimes n+k}) \stackrel r \rightarrow H^0(Y,
L^{\otimes n+k} |_Y) \rightarrow H^1(L^{\otimes k})
\rightarrow H^1(L^{\otimes n+k}) \rightarrow 0.
$$
The arrow $r$ of this sequence is the restriction map;
if it vanishes, we have $\dim H^0(Y, L^{\otimes n+k}|_Y) \leq
\dim H^1(L^{\otimes k})$ for all $n\gg 0$ and $k$.
The last term of this exact sequence actually
implies that $\dim H^1(L^{\otimes n+k})\leq \dim H^1(L^{\otimes k})$,
for all $k$, and all $n\gg 0$,
hence $H^1(L^{\otimes n+k})$ is bounded by a universal constant.
This implies that $\dim H^0(Y, L^{\otimes n+k} |_Y)$
is also bounded, whenever $r=0$, which is
impossible, because $L|_Y$ is ample.