Noting that $P_\ell(x)$ is even for even $\ell$ and odd for odd $\ell$, we have the relation
$\int_{-1}^0 P_\ell(x)\mathrm dx=(-1)^{\ell}\int_0^1 P_\ell(x)\mathrm dx$
so we can concentrate on evaluating
$\int_0^1 P_\ell(x)\mathrm dx$
for which we can use the integral expression
$\int P_\ell(x)\mathrm dx=\frac{P_{\ell+1}(x)-P_{\ell-1}(x)}{2\ell+1}$
and the special values $P_\ell(1)=1$ (due to normalization) and
$P_\ell(0)=\begin{cases}\frac{(-1)^{\ell/2}}{2^\ell}\binom{\ell}{\ell/2}&\ell\;\text{even}\\0&\ell\;\text{odd}\end{cases}$
For even $\ell$, it is easy to see from these considerations that the integral is equal to 1 for $\ell=0$, and 0 otherwise. This leaves us the case of odd $\ell$, where we obtain the expression
$\int_0^1 P_{2\nu+1}(x)\mathrm dx=\frac{(-1)^\nu}{2^{2\nu+1}(\nu+1)}\binom{2\nu}{\nu}$
The cheap-ass way of determining the value of $\int_0^1 P_{2\nu}(x)\mathrm dx$ without going through the rigamarole above is to note that since $P_{2\mu}(x)$ and $P_{2\nu}(x)$ for integer $\mu$ and $\nu$ are both even functions, their product is also an even function, and thus
$\int_0^1 P_{2\mu}(x)P_{2\nu}(x)\mathrm dx=\frac12\int_{-1}^1 P_{2\mu}(x)P_{2\nu}(x)\mathrm dx$
after which, we then recall the orthogonality relation for the Legendre polynomials
$\int_{-1}^1 P_{2\mu}(x)P_{2\nu}(x)\mathrm dx=\frac{2\delta_{2\mu,2\nu}}{4\mu+1}$
and then consider letting $\nu=0$...