Plus/Minus Model accuracy from $R^2$ I completed a linear regression for a model I was working on, and obtained that the $R^2$ value was $R^2 = 0.801$. 
Can one assess a $\pm$ error from this value for future predictions? I.e., if I now use this linear model to predict against a new set of data, can I use this $R^2$ value to get a  $\pm$ value on that prediction? 
 A: No, you can't, for two reasons.


*

*$R^2$ indicates the proportion of variance explained by your model. $R^2=0.80$ can mean that you explain 80% of very little variance, so your prediction-interval (PI) should be small. Or it can mean that you explain 80% of a huge lot of variance, so your PIs should be large.

*$R^2$ is an in-sample measure of model fit. In-sample fits are very misleading as guides to out-of-sample predictive accuracy.
To calculate PIs for multiple regression, take a look at this earlier thread: How to calculate the prediction interval for an OLS multiple regression?
A: You are talking about Root Mean Squared Error of Prediction (RMSEP).  
It is fundamentally different than the $R^2$ value, and they are not related in the way you are hoping.  Your $R^2$ value is the approximate amount of $y$-variance (dependent variable variance) explained by your $x$-matrix of covariates (independent variables).  This is answering the question "how much variance can I explain with my given set of predictors?"
Your RMSEP (explained in this website on calibration) is the approximate error that your model will produce in predicting a future out-of-sample value.  It is a metric used to answer "if I use my current model in the real-world, how much error will it produce while predicting?" 
$R^2$ is used to determine how much variance a model explains.  RMSEP is used to determine how well your model can predict out-of-sample values.  They are not related.
