I think the minimum is $l=\frac{5^{m+1}-1}{4}$. Also, if a value $l$ works, then $5l$ must also work. I think you can express $A=\frac{(4(5^m+1)l)!}{(4l)!}$, then use the fact that: $$v_p(n!)=\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\cdots$$

Since $V_{p}(m!)=\frac{m-s_{p}(m)}{p-1}$ , in which $s_{p}(n)$ meaning the sum of the digits of $n$ in the base of $p$, we have \begin{align*} V_{5}(A_{l})&=\frac{4*5^{m} *l +4l - s_{5}(4*5^{m} *l +4l) - 4l + s_{5}(4l) } {4}\\ &=\frac{4*5^{m} *l - s_{5}(4*5^{m} *l +4l) + s_{5}(4l) } {4}\\ &=4*5^{m} *l - \frac{s_{5}(4*5^{m} *l +4l) - s_{5}(4l) } {4}\\ \end{align*}Therefore the question changed to be: finding infinite $l$ such that $s_{5}(4*5^{m} *l +4l) = s_{5}(4l)$ That means, if we take $l=\frac{5^{k}-1}{4}$ which $k \geq m+1$ we will have: $s_{5}(4l)=s_{5}(5^{k}-1)=4k$ \begin{align*} s_{5}(4*5^{m} *l +4l) &= s_{5}(5^{m+k}-5^{m}+5^{k}-1)\\ &= s_{5}(5^{m+k}+(5^{k}-5^{m}-1))\\ &= 1+ 4k-1 =4k\\ \end{align*}That proofs that part one. In the part 2, we need to proof that $l=\frac{5^{m+1}-1}{4}$ is the smallest number satisfying the condition. First we count how many digit we need to have in the number of $4l$ in counting in the base of 5. Since we need to let $s_{5}(4*5^{m} *l +4l) = s_{5}(4l)$, we need at least $m+1$ digits or $s_{5}(4*5^{m} *l +4l) = 2*s_{5}(4l)$, and also ${5^{m+1}-1}$ are having the exact $m+1$ digits, Therefore we can assume that $4l=\displaystyle\sum\limits_{i=0}^m c_{i} 5^{i}$, in which $c_i \in \left\{ 0,1,2,3,4 \right\}$ and $c_{m} \neq 0$ Thus, \begin{align*} & s_{5}(4l) = \displaystyle\sum\limits_{i=0}^m c_{i}\\ & s_{5}(4*5^{m} *l +4l) = s_{5}(\displaystyle\sum\limits_{i=0}^{m-1} c_{i} 5^{i} + (c_0 + c_m) 5^m + \displaystyle\sum\limits_{i=1}^{m} c_{i} 5^{m+i}) > \displaystyle\sum\limits_{i=0}^m c_{i}\\ \end{align*}if $c_m \neq 4$ Therefore $c_m =4$ and that means $c_i =4$ for any $i=0,1,2, ... ,n$ That means $4l=5^{m+1}-1$ and $l=\frac{5^{m+1}-1}{4}$ , Q.E.D.

\begin{align*} v_5(A)&=v_5((4(5^m+1)l)!)-v_5((4l)!)\\ &=\left(\left\lfloor\frac{4(5^m+1)l}{5}\right\rfloor+\left\lfloor\frac{4(5^m+1)l}{5^2}\right\rfloor+\cdots+\left\lfloor\frac{4(5^m+1)l}{5^m}\right\rfloor+\left\lfloor\frac{4(5^m+1)l}{5^{m+1}}\right\rfloor+\left\lfloor\frac{4(5^m+1)l}{5^{m+2}}\right\rfloor+\cdots\right)\\ &-\left(\left\lfloor\frac{4l}{5}\right\rfloor+\left\lfloor\frac{4l}{5^2}\right\rfloor+\left\lfloor\frac{4l}{5^3}\right\rfloor+\cdots\right)\\ &=\left(4\cdot5^{m-1}l+4\cdot5^{m-2}+\cdots+4+\left\lfloor\frac{4l}{5}\right\rfloor+\left\lfloor\frac{4l}{5^2}\right\rfloor+\cdots+\left\lfloor\frac{4l}{5^m}\right\rfloor+\left\lfloor\frac{4l}{5}+\frac{4l}{5^{m+1}}\right\rfloor+\left\lfloor\frac{4l}{5^2}+\frac{4l}{5^{m+2}}\right\rfloor+\cdots\right)\\ &-\left(\left\lfloor\frac{4l}{5}\right\rfloor+\left\lfloor\frac{4l}{5^2}\right\rfloor+\left\lfloor\frac{4l}{5^3}\right\rfloor+\cdots\right)\\ &=5^ml-l+\left(\left\lfloor\frac{4l}{5}+\frac{4l}{5^{m+1}}\right\rfloor+\left\lfloor\frac{4l}{5^2}+\frac{4l}{5^{m+2}}\right\rfloor+\cdots\right)-\left(\left\lfloor\frac{4l}{5^{m+1}}\right\rfloor+\left\lfloor\frac{4l}{5^{m+2}}\right\rfloor+\cdots\right) \end{align*} Therefore, we must find a solution to: $$\left\lfloor\frac{4l}{5}+\frac{4l}{5^{m+1}}\right\rfloor+\left\lfloor\frac{4l}{5^2}+\frac{4l}{5^{m+2}}\right\rfloor+\cdots=l+\left\lfloor\frac{4l}{5^{m+1}}\right\rfloor+\left\lfloor\frac{4l}{5^{m+2}}\right\rfloor+\cdots$$

Let $s(n)$ be the sum of digit of $n$ is base $5$. Recall that $\nu_5(n!) = \frac{n-s(n)}{4}$ thus the condition rearranges to $$\nu_5([4(5^m+1)l]!) - \nu_5((4l)!) = 5^ml \iff s(4(5^m+1)l) = s(4l).$$Observe that if $l$ works, then $5l$ works so the second part would automatically prove the first part. Now we claim that $\frac{5^{m+1}-1}{4}$ is the minimum. It works because \begin{align*} s(4(5^m+1)l) &= s((5^m+1)(5^{m+1}-1))\\ &= s(1\underbrace{00...0}_{m}3\underbrace{44...4}_{m})\\\ &= 4(m+1)\end{align*}But clearly $s(4l) = 4(m+1)$ thus $\frac{5^{m+1}-1}{4}$ works as claimed. In particular, this completes the first part. Now we proceed to prove that this value is minimum. Suppose that a smaller $l$ works. Let $k=4l$. As $k < 5^{m+1}$, we can write $$k = a_m5^m + a_{m-1}5^{m-1} +...+a_1\cdot 5 + a_0\quad (a_0,a_1,...,a_m\in\{0,1,2,3,4\}).$$Thus $$s((5^m+1)k) = 5^m(k+a_m) + a_{m-1}5^{m-1} + a_{m-2}5^{m-2} + ... + a_1\cdot 5 + a_0$$Thus $s((5^m+1)k) = s(k)$ if and only if $s(k+a_m) = a_m$. Now, the key observation is during adding $k$ to $a_m$, there is a carry to the leftmost digit. However, if the carry change $a_m$ to $a_m+1$, then it would be impossible to have $s(k+a_m)=a_m$. Thus there is a carry from $5^m$ digit to $5^{m+1}$ digit. But $a_m$ is a digit, so a carry must occur in every digit, including a carry from the $5^m$ digit to the $5^{m+1}$ digit. Thus there are $m+1$ carries, implying $$\frac{s(k) + s(a_m) - s(k+a_m)}{4} = m+1\implies s(k) = 4(m+1)$$which implies $k=5^{m+1}-1$, which is not smaller, contradiction.