How long could Earth's oxygen supply last if no new oxygen were produced?
Disclaimer: 

This is going to be a very mathematical answer. Before answering it, I assume that you are only asking about humans, assuming that all other organisms don't require $O_2$ to survive (as it will complicate the answer many many times).

List of variables:

V     = total volume of air on earth (in l)
V'    = total volume of oxygen on earth (in l)
V(1)  = volume of oxygen used by 1 person in whole lifetime (in l)
V(a)  = volume of oxygen used by whole mankind (one generation) in whole lifetime (in l)
V(p)  = percentage of total oxygen used by whole mankind (one generation) in whole lifetime


Method used:

$ \%\hspace{1mm}Volume\hspace{1mm}of\hspace{1mm}O_2\hspace{1mm}used\hspace{1mm}per\hspace{1mm}generation = \underline{\hspace{3mm}Volume\hspace{5mm}of\hspace{5mm}O_2\hspace{5mm}in\hspace{5mm}air\hspace{3mm}}\times\hspace{1mm}100$
$\hspace{83mm}Volume\hspace{1mm}of\hspace{1mm}O_2\hspace{1mm}used\hspace{1mm}by\hspace{1mm}humans$

$Number\hspace{1mm}of\hspace{1mm}years = \underline{\hspace{2mm}Number\hspace{1mm}of\hspace{1mm}years\hspace{1mm}in\hspace{1mm}1\hspace{1mm}generation\hspace{2mm}}$
$\hspace{42mm}\%\hspace{1mm}Volume\hspace{1mm}of\hspace{1mm}O_2\hspace{1mm}used\hspace{1mm}per\hspace{1mm}generation$

Procedure:

Lets first find out the total amount of air on earth.

From this:


  The air sphere measures 1999 kilometres across and weighs 5140 trillion tonnes. 


From here:


  At sea level and at 15 °C air has a density of approximately 1.225 kg/m3


Total volume of air:

$volume = mass \hspace{1mm} / \hspace{1mm} density$

$V = 5140 \times 10^{12} \hspace{1mm} tonnes \hspace{1mm} / \hspace{1mm} 1.225 \hspace{1mm} kg.m^{-3}$

$V = 4195.9 \times 10^3 \times 10^{12} \hspace{1mm} m^3$

$V = 4195.9 \times 10^3 \times 10^{12} \times 10^3 \hspace{1mm} l$

$V = 4195.9 \times 10^{18} \hspace{1mm} l$

Now, from this, 20.95% of this is $O_2$ i.e.

$V' = 4195.9 \times 10^{18} \times 20.95 / 100$

$V' = 87904.489 \times 10^{16}$

$V' = 879.04 \times 10^{18} \hspace{1mm} l$

Moving on to next part, from this:


  a human being uses about 550 liters of pure oxygen per day.


From this:


  71 years was the average life expectancy at birth of the global population in 2013.


$V_1 = 550 \hspace{1mm} l.day^{-1}.person^{-1} \times 25915 \hspace{1mm} days$

$V_1= 142.53 \times 10^5 \hspace{1mm} l.person^{-1}$

From this:


  The world population (the total number of living humans on Earth) was 7.349 billion as of July 1, 2015.


So

$V_{a} = 142.53 \times 10^5 \hspace{1mm} l.person^{-1} \times 7.349 \times 10^9 people$

$V_{a} = 1047.452 \times 10^{14} \hspace{1mm} l$

Finally

$V_p = V_a / V' \times 100$

$V_p = (1047.452 \times 10^{14}) / (879.04 \times 10^{18}) \times 100$

$V_p = 1.191 \times 10^{-2}$

$V_p = 0.0191 \hspace{1mm} \% $

This gives the percentage of total $O_2$ in air used by one human generation. From this, we can find the number of years for which $O_2$ will last, as I have done below:

$25,915 \hspace{1mm} days = 71 \hspace{1mm} years$

$71 \hspace{1mm} years = 0.0191 \hspace{1mm} \%$

$1 \hspace{1mm} \% = 71 / 0.0191 \hspace{1mm} years$

$100 \hspace{1mm} \% = 3717.277 \times 100 \hspace{1mm} years$

$\hspace{13mm} = 371,727.748 \hspace{1mm} years$

Conclusion:

So, yes at least humans can live for a very long time without plants doing any photosynthesis (ideally). 371,727 years seems too large a period of time, but it is just based on ideal assumptions, this number would come down (maybe into your specified range) when more variables are added.

For example:

To show extent of deviation by addition of factors, I add a very general, but complex factor in the expression: air pollution.

How much $CO_2$ is released per year can be estimated from this:


  In 2011, utility coal plants in the United States emitted a total of 1.7 billion tons of CO2


And with the contribution of countries like China, India, Europe, etc. that number should rise up to at least 2.5 million tonnes (its just that I couldn't find the worldwide data on it).

Now:

$1 \hspace{1mm} C + 1 \hspace{1mm} O_2 \rightarrow 1 \hspace{1mm} CO_2$

$2.5 \times 10^9 \hspace{1mm} kg \hspace{1mm} CO_2 \Rightarrow 2.5 \times 10^9 \hspace{1mm} kg \hspace{1mm} O_2$

So, 2.5 million tonnes more $O_2$ is being used each year from the environment. But, there is another twist in this. From this:


  WHO reports that in 2012 around 7 million people died - one in eight of total global deaths – as a result of air pollution exposure.


It means that 7 million people less will consume $O_2$ each year i.e.

$Net \hspace{1mm} population = 7349 \times 10^6 - 7 \times 10^6$

$\hspace{32mm} = 7.342 \times 10^9$

$V_{a} = 142.53 \times 10^5 \times 7.342 \times 10^9$

$V_{a} = 1046.455 \times 10^{14} \hspace{1mm} l$

$V_p = V_a / V' \times 100$

$V_p = (1046.455 \times 10^{14}) / (879.04 \times 10^{18}) \times 100$

$V_p = 1.1904 \times 10^{-2}$

$V_p = 0.01904 \hspace{1mm} \% $

$25,915 \hspace{1mm} days = 71 \hspace{1mm} years$

$71 \hspace{1mm} years = 0.01904 \hspace{1mm} \%$

$1 \hspace{1mm} \% = 71 / 0.01904 \hspace{1mm} years$

$100 \hspace{1mm} \% = 3728.991 \times 100 \hspace{1mm} years$

$\hspace{13mm} = 372,899.159 \hspace{1mm} years$

So, overall number increases instead of decreasing. This example was just to show how adding more factors to the equation would deviate the overall answer, and how large that deviation can be.