I have a matrix, say $A$, which is square of dimension $n$ and symmetric by construction. The dimension $n$ is large, i.e. it is allowed to pass to infinity.
I can show that, for every $n$-dimensional vector $b$ which is bounded in $L_{1}$-norm, i.e. $\sum_{i=1}^{n}|b_{i}| < \infty$, we have that
$b'Ab>0$,
so in this sense $A$ is positive definite.
Can I say anything about the smallest eigenvalue of $A$ with this information? All I would want to know is whether the smallest eigenvalue is positive, I do not need any "sophisticated"/sharp bounds.
If $\lambda$ is the smallest eigenvalue of $A$, then it has an eigenvector $b$. So $Ab = \lambda b$ and $b'Ab = \lambda b'b$. Since both $b'Ab > 0$ (as Giuseppi Negro pointed out, your $L^1$ condition holds for all $b$) and $b'b > 0$, it must be that $\lambda > 0$.