The security of $\{0,1\}$ error LWE depends heavily on the number of samples you get access to.
By this, for a standard "LWE sample"
$$(A, As + e)$$
where $A\in\mathbb{Z}_q^{m\times n}$, $s\in\mathbb{Z}_q^n$, $e\in\mathbb{Z}_q^m$, I mean the parameter $m$.
This heavy dependence is because for $m = \Omega(n^2)$ there are polynomial-time attacks, i.e. it is completely insecure.
Alternatively, things are thought to be ok for $m = O(n)$.
From this, we see that
(Heuristically) for any $\epsilon > 0$, binary error LWE can be solved in polynomial time $n^{O(1/\epsilon)}$ given $\epsilon\cdot n^2$ samples, and for $0 < \alpha < 1$ it can be solved in sub-exponential time $2^{O(n^{1-\alpha})}$ time given $n^{1+\alpha}$ samples.
This is for algebraically unstructured LWE.
For algebraically structured LWE, weaker results are known (see this, though I will not summarize it here as it is quite technical).
You also asked about choosing the secret from $\{0,1\}$.
This is much better supported theoretically, though there are some caveats
- For (plain) LWE (i.e. algebraically unstructured), there is no security loss provided you increase the dimension to componsate, i.e. use dimension $n\log_2 q$ rather than dimension $n$
- For algebraically structured LWE, there is plausibly a security loss. Still, using ternary secrets $\{-1,0,1\}$ is quite common, and was standardized in the Homomorphic Encryption standard (binary secrets were not).
