Step 1: Identify the parabola.
Compare $y^2=8x$ with $y^2=4ax$. So $4a=8$ and $a=2$.
Step 2: Use parametric points.
A point on the parabola is $P(at^2,2at)=(2t^2,4t)$. The tangent at $P$ is $ty=x+at^2$, that is $x-ty+2t^2=0$.
Step 3: Find $OQ$, the foot of the perpendicular from $O$.
$Q$ is where the perpendicular from the origin meets the tangent, so $OQ$ is the distance from the origin to the tangent: \[ OQ=\frac{|2t^2|}{\sqrt{1+t^2}}. \]
Step 4: Find $OP$.
\[ OP=\sqrt{(2t^2)^2+(4t)^2}=\sqrt{4t^4+16t^2}=2t\sqrt{t^2+4}. \]
Step 5: Use the standard result for this set up.
For this perpendicular-from-vertex configuration on $y^2=4ax$, the product $OP\cdot OQ=4a^2$. The lengths combine so the variable $t$ cancels, leaving only $a$.
Step 6: Put in $a=2$.
\[ OP\cdot OQ=4(2)^2=16. \] \[ \boxed{16} \]