Why aren't +/- test leads connected to battery terminals attracted to each other? Similar to static electricity, why wouldn't the negative charges accumulate on the wire connected to the negative terminal of a battery be attracted to the positive wire (lead) when their ends are brought close to each other?  Is there an attractive force but it is just too weak to be felt by hand?  What if a dielectric material connected to both ends of the leads; would that change anything?
 A: They are attracted to each other, the force is just too small to feel it. We can estimate it as follows: 
The force between two charged plates is $F=E∗Q/2$, where $E$ is the electric field between plates. We can express this by the voltage between the plates $U$ and plate distance $d$ and get $F=U∗Q/2d$. Finally we substitute $Q=C*U$ for the charge as a function of voltage and capacitance $C$ and get $F=U^2*C/2d$. 
If we insert $U=12Volt$ for a typical battery voltage, $C=100pF$ for the capacitance between the lead wires and $d=1mm$ for the smallest distance between them, then we get $F=7.2μN$ for the force, which is quite small. I was, of course, cheating a bit because I just gave you a random estimate for the capacitance of the wires, but it is in the right ballpark. 
Now, even if we raise the voltage to $120V$, the force would be no more than $720μN$. If you could go to $2500V$ without electrocuting yourself (don't!), the force may be barely perceptible. That's kind of the voltage range in which electrostatic speakers operate, which is one of the few applications which use electrostatic forces directly to generate movement. 
