Let $|z+4|\le 3.$ This represents all complex numbers $z$ whose distance from the point $-4$ on the Argand plane is at most $3$.

We need to find the maximum value of $|z+1|,$ which is the distance of $z$ from the point $-1$.

So the problem is geometric:

• Center of the given disk: $-4$
• Radius of the disk: $3$
• Point from which distance is measured: $-1$

The distance between the center $-4$ and the point $-1$ is $|-4-(-1)|=|-3|=3.$

The maximum distance from $-1$ to any point in the disk centered at $-4$ with radius $3$ is $3+3=6.$ This follows from the triangle inequality, and equality is attained when $z$ lies on the boundary in the direction away from $-1$.

Algebraically, using triangle inequality: $|z+1|=|(z+4)-3|\le |z+4|+3\le 3+3=6.$ Hence, $|z+1|_{\max}=6.$

Therefore, the correct option is: $\boxed{A}$