FullSimplify[(1/(2^
       j) ((k*Gamma[5 + 2 j] Gamma[
          1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2, 
           7/2 + j + l/2}, -1])/
       Gamma[6 + 2 j + 
         l] + ((k + m) Gamma[7 + 2 j] Gamma[
          1 + l] HypergeometricPFQ[{1, 7/2 + j, 4 + j}, {4 + j + l/2, 
           9/2 + j + l/2}, -1])/Gamma[8 + 2 j + l]))/(2^(-5 - 3 j - 
       l) Gamma[5 + 2 j] Gamma[
     1 + l] (k HypergeometricPFQRegularized[{1, 5/2 + j, 
         3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1] + 
      1/2 (3 + j) (5 + 2 j) (k + m) HypergeometricPFQRegularized[{1, 
         7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1])),
 {j \[Element] Integers, k \[Element] Integers, 
  l \[Element] Integers, m \[Element] Integers}]

returns Sqrt[Pi], so your hypothesis seems to be confirmed by Mathematica. I'm not sure it's easy to produce a sequence of simplifying steps to "prove" that for humans.

5-10-2023
Is this an approach for irrational roots of transcendental number estimations?
It is interesting also because one always wonders if we have accounted for all numbers in the continuum.
There have been so many surprises over the millennia.

When ever something, you write down, evaluates to Sqrt[Pi] it comes from an integral over a full circle squared. Since you have Gamma functions, there is always a

Sqrt[Pi] =Gamma[1/2]

around.

Best example`

(Integrate[Exp[-x^2],{x,-Infinity, Infinity}])^2 == Pi

This can be demonstrated by the fact, that the product of two integrals equals the Integral of the product with two different variables

Integrate[Exp[-x^2-y^2]  , {x,-oo,oo}, {y,-oo,oo}] == 
  Integrate[Exp[-r^2] r {r, 0, oo}] 2 Pi  ==  Integrate[Exp[-u] ,{u, 0, oo}] Pi

If you find an identity for Pi or sqrt Pi, its an identiy in the integers. where the transcendent is as factor, mostly stemming from integrals over circles in the complex plane.