Calculating the biological standard potential
Unfortunately, the OP's Nernst equation is incorrect for several reasons. The balanced net redox equation is

$$\ce{5 Sn^{4+} (aq) + 2 Mn^{2+} (aq) + 8 H2O &lt;=&gt; 5 Sn^{2+} (aq) + 2 MnO4^- (aq) + 16 H+ (aq) \tag 1}$$
and n = 10 electrons for the reaction as written. The reaction quotient, Q, is

$$\ce{Q = \frac{[Sn^{2+}]^5 [MnO4^-]^2 [H+]^16}{[Sn^{4+}]^5 [Mn^{2+}]^2}              \tag 2}$$

and the Nernst equation is

$$\displaystyle{E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln(Q) = E_{cell}^\circ - \frac{RTln(10)}{nF} \log(Q)         \tag 3}    $$

The OP's question states that "all species except $\ce{H+}$ are in their standard states, so their activities are all equal to 1". Hence, in this very contrived example, Q reduces dramatically:

$$\displaystyle{Q = [H^+]^{16}             \tag 4}$$

The OP notes that $\displaystyle{E_{cell}^\circ }$ = 1.36 V (from tables). With T = 298.15 K, $R = \pu{8.31446261815324 J mol-1 K-1}$ and $F = \pu{96485.3321233100184 C mol-1}$, then

$$\ce{\displaystyle{ \frac{RTln(10)}{nF} \approx \frac{0.05916 \ V}{n} }        \tag 5}$$

So the Nernst equation is simply

$$\displaystyle{E_{cell} \approx 1.36 \ V - \frac{0.05916 \ V}{n} \log(Q) = 1.36 \ V - \frac{0.05916 \ V}{10} \log([H^+]^{16})      \tag 6}    $$

Hence, using $\ce{pH = -log[H^+] = +log(1/[H^+])}$, the Nernst equation can be expressed as

$$\displaystyle{E_{cell} \approx 1.36 \ V - 16 \times \frac{0.05916 \ V}{10} \log[H^+] =  1.36 \ V + 16 \times \frac{0.05916 \ V}{10} pH      \tag 7}    $$

If pH = 7, then $\ce{E_{cell} \approx 2.02 \ V}.$

It is better to use expression (3) in order to avoid getting too tricky. This is where the OP went wrong: forgetting the "16" exponent on the hydrogen ion concentration and also missing the fact that the negative sign before the pre-log correction term must be subsumed into the pH definition. In a realistic example, it will not happen that everything is in standard state except the hydrogen ion.

What about dividing the net ionic equation, i.e., equation (1), by 16? Then n = 10/16 = 5/8 and all the exponents in the Q expression also get divided by 16. This is no problem if everything is in standard state except the hydrogen ion. Otherwise, it is just facilitating making a silly mistake.