Should the lambda for L1 norm regularizer inversely be proportional to the number of trainable weights? Say I want to implement Conv2D in keras and for each Conv2D layer, if I apply 20 filters of [2,3] filter on an input with depth of 10, then there will be 20*(2*3*10+1) = 1220 trainable weights. 

the value of L1 norm would proportionally increase the more trainable weights there are. Similarly for L2 norm. 

So shouldn't the lambda, as in kernel_regularizer=l1(lambda), be inversely be proportional to the number of trainable weights? 

intuitively for me, if lambda of 0.1 worked for 10,000 weights, then applying the same or bigger lambda for 1 million weights doesn't make sense to me.
Backpropagation doesn't handle Regularization like this i.e. if you are thinking "10 weights make penalty 100, so 100 weights will make penalty 1000. 
So let's have a smaller $\lambda$

Backpropagation uses the partial differentiation of the Loss. Now the Loss has an extra $\Sigma$$w_i^2$ and so the derivative will have an extra piece which will be proportional to the weight(Derivative of $x^2$ = $2*x$). 
It will be separate for every weight, so the number of weights will have no impact

The new equation will be - 
$w_i$ = $w_i$ - (backprop stuff as before) - $\lambda$*$w_i$

$\lambda$ is just a knob. You also have any value proportional to $w_i$ e.g. 5% of $w$.