You ask whether an unconditional basis must be a Schauder basis. The answer is no even when $V$ is a separable normed space. Let $B$ be an orthonormal basis for $L_2(0,1)$ so that no element of $B$ is in $L_\infty$. Then $B$ is an unconditional basis for the incomplete space $L_2(0,1)$ with the $L_1$ norm, but none of the coordinate functionals are continuous.

EDIT 10/22/10. Consider $f(t) = |t|^{-1/4}$ in $L_2(-\pi,\pi)$ with normalized Lebesgue measure. Gram-Schmidt $f, e^{ix}, e^{-ix}, e^{2ix}, e^{-2ix}, ,,,$; this produces an orthonormal basis $g_n$ for $L_2(-\pi,\pi)$. Since the Fourier coefficients of $f$ are all non zero, no $g_n$ is in $L_\infty$. Since successive Fourier coefficients of $f$ are comparable (they decay like $n^{-3/4}$), the $L_1$ norms $\|g_n\|_1$ are bounded away from zero.

Suppose that $\sum a_n g_n$ unconditionally converges in $L_1$ to zero. Since $L_1$ has cotype two, $\sum \|a_n g_n\|_1^2 <\infty$, so that
$\sum \|a_n|^2 <\infty$. Thus $\sum a_n g_n$ converges in $L_2$ and the sum can be zero iff $a_n=0$ for all $n$.

I do not see that $g_n$ is a basis for $(L_2, \| \cdot\|_1)$. That is, if
$\sum_{n=1}^\infty a_n g_n$ converges to zero in $L_1$, must $a_n=0$ for all $n$?