Gibbs free energy and its change for the hydrolysis of ATP
You can use tables of standard transformed Gibbs energies of formation, $\ce{\Delta _fG^{'o}}$ , such as those found in Table 1 of the following reference, to calculate apparent equilibrium constants of biochemical reactions, and to calculate standard transformed Gibbs energies of formation, $\ce{\Delta _rG^{'o}}$ (at pH 7), which I think is exactly what you require. 


Calculating apparent equilibrium constants of enzyme-catalyzed
reactions at pH 7 (by R. A. Alberty)


The first thing to say it that you need to include water as a reactant. 

$$\ce{ATP + H2O = ADP + P_i}\tag{1}$$

Secondly, the reaction shown in Eqn(1) is used as an example by Alberty in the paper cited above, where he calculates both $\ce{\Delta _rG^{'o}}$ (at pH 7) and the apparent equilibrium constant (at pH 7). 

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The following values were taken from Table 1 of the Alberty reference cited above.  They refer to pH 7, 25oC, and an ionic strength of 0.1. 



&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;$\ce{\Delta _fG^{'o}/{kJ mol^{-1}}}$

ATP             -2276.43

ADP             -1409.95

Pi              -1059.17

H2O              -156.05



Knowing these values, it is now very easy to calculate the standard transformed Gibbs energies of formation, $\ce{\Delta _rG^{'o}}$, for the reaction shown in Eqn (1) (it is merely a case of 'pluggin in' the values). 

$$\Delta _rG^{\ce{'o}} = \sum v_i^{'} \Delta_f G_i{^o} = (2276.43 + 156.05 - 1409.95 - 1059.17)\tag{2}$$

$$\ce{\Delta _rG^{'o} = -36.68 kJ mol^{-1}\tag{3}}$$

In Eqn (2), $v_i^{'}$ is the stoichiometric number (positive for reactants
and negative for products)

The criterion which determines whether a reaction will 'go' in the forward direction or the reverse direction is the transformed Gibbs energy of reaction, $\Delta_rG^{'}$, (not $\Delta_rG^{'o}$). 


  If  $\Delta_rG^{'}$ is negative, the reaction goes in the forward direction. 


We can say that under standard conditions, Eqn (1) goes to the right, and as $\Delta_rG^{'o}$ (not $\Delta_rG^{'}$) is directly related to the apparent equilibrium constant (Eqn 4), we can say that the equilibrium constant for the rection of Eqn (1) is very much to the right. 

$$\Delta _rG^{\ce{'o}} = \ce{-RT ln K^{'}}\tag{4}$$

We can go a step further (Alberty,2000) and calculate the apparent equilibrium constant, $K^{'}$, for the reaction shown in Eqn (1). 

$$K^{'} = e^{\frac{-\Delta _rG^{'o}}{RT}}=e^{36680/(8.3145 \times 298.15)}\tag{5}$$

$$K^{'}  = 2.67 \times 10^6\tag{6}$$

The great value of tables of $\ce{\Delta _fG^{'o}}$ it that they allow $\Delta _rG^{\ce{'o}}$ and $K^{'}$ to be calculated for many reactions, even those for which there is no known enzyme, or where the equilibrium constant has not been determined. 

Notes

In Eqn (5) above, $R$ is the gas constant ($\ce{8.3145 J K^{-1} mol^{-1}}$
) and $T$ is the temperature (in Kelvin)

IMO, the Alberty reference cited above is a model of clarity that makes these type of calculations very easy. It is also an 'easy read' with many great examples. 

Two other good refs, if you are 'into' such things, are 


Biochemical thermodynamics (by R.A. Alberty)
Recommendations for nomenclature and tables in biochemical thermodynamics (by R.A. Alberty)


Other  tables of standard transformed Gibbs energies of formation, $\ce{\Delta _fG^{'o}}$, may be found in 


Calculation of Standard Transformed Gibbs Energies and Standard Transformed Enthalpies of Biochemical Reactants  (by R.A. Alberty)
Calculation  of Standard Transformed Formation Properties of Biochemical Reactants and Standard Apparent Reduction Potentials of Half Reactions (by R.A. Alberty)